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What stops an assembly program from crashing the operating system?
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$begingroup$
First of all, I am a beginner, so if this question sounds silly, please point out the incorrect assumptions.
From what I understand, the job of an operating system is to manage hardware, and the software that runs on the OS. Also, from what I understand, assembly programs allows one to control hardware almost directly. In an assembly program, one can read and write data into registers, and read and write data into RAM.
Given this freedom to mess with registers and RAM, would it not be possible for assembly programs to affect the operating system? Suppose the OS is using register A to store crucial information. Suppose I write an assembly program, assemble the program, and run the assembled program on the OS. If the program successfully writes junk into register A, the OS will surely be affected.
Questions:
Is it possible to mess with register A in the manner described above?
If not, what stops assembly programs from modifying the registers used by the OS?
assembly register operating-system
$endgroup$
|
show 2 more comments
$begingroup$
First of all, I am a beginner, so if this question sounds silly, please point out the incorrect assumptions.
From what I understand, the job of an operating system is to manage hardware, and the software that runs on the OS. Also, from what I understand, assembly programs allows one to control hardware almost directly. In an assembly program, one can read and write data into registers, and read and write data into RAM.
Given this freedom to mess with registers and RAM, would it not be possible for assembly programs to affect the operating system? Suppose the OS is using register A to store crucial information. Suppose I write an assembly program, assemble the program, and run the assembled program on the OS. If the program successfully writes junk into register A, the OS will surely be affected.
Questions:
Is it possible to mess with register A in the manner described above?
If not, what stops assembly programs from modifying the registers used by the OS?
assembly register operating-system
$endgroup$
1
$begingroup$
smart programmers...
$endgroup$
– Sunnyskyguy EE75
1 hour ago
$begingroup$
There are a lot of computer architectures now and were in the past, as well as many operation systems were developed. What architecture/OS are you referring exactly? On some (old) architectures there were no possibility to stop program from what it was doing after start, that's correct. But modern hardware/OS have built in hardware tools which give only the part of the memory for the program in "normal" mode (not superuser), and it can't access memory outside this limit. Registers are free to use, as OS don't store any useful info in registers, just in memory/on disk.
$endgroup$
– cyclone125
1 hour ago
$begingroup$
In a microprocessor your program runs in "user mode", the operating systems runs in "system mode." If a user mode program executed a halt instruction, for example, the machine would not halt. The halt would be trapped and the operating system invoked. Regarding RAM, the OS would set up an environment for the user mode program so that, via memory management hardware, what the user program sees as RAM address X would not really be RAM address X.
$endgroup$
– George White
1 hour ago
$begingroup$
@cyclone125 I am not referring to any architecture/OS in particular. It's just a general question.
$endgroup$
– Flux
1 hour ago
1
$begingroup$
... Long ago I used to write in raw machine code. Yee ha !!! :-)
$endgroup$
– Russell McMahon
56 mins ago
|
show 2 more comments
$begingroup$
First of all, I am a beginner, so if this question sounds silly, please point out the incorrect assumptions.
From what I understand, the job of an operating system is to manage hardware, and the software that runs on the OS. Also, from what I understand, assembly programs allows one to control hardware almost directly. In an assembly program, one can read and write data into registers, and read and write data into RAM.
Given this freedom to mess with registers and RAM, would it not be possible for assembly programs to affect the operating system? Suppose the OS is using register A to store crucial information. Suppose I write an assembly program, assemble the program, and run the assembled program on the OS. If the program successfully writes junk into register A, the OS will surely be affected.
Questions:
Is it possible to mess with register A in the manner described above?
If not, what stops assembly programs from modifying the registers used by the OS?
assembly register operating-system
$endgroup$
First of all, I am a beginner, so if this question sounds silly, please point out the incorrect assumptions.
From what I understand, the job of an operating system is to manage hardware, and the software that runs on the OS. Also, from what I understand, assembly programs allows one to control hardware almost directly. In an assembly program, one can read and write data into registers, and read and write data into RAM.
Given this freedom to mess with registers and RAM, would it not be possible for assembly programs to affect the operating system? Suppose the OS is using register A to store crucial information. Suppose I write an assembly program, assemble the program, and run the assembled program on the OS. If the program successfully writes junk into register A, the OS will surely be affected.
Questions:
Is it possible to mess with register A in the manner described above?
If not, what stops assembly programs from modifying the registers used by the OS?
assembly register operating-system
assembly register operating-system
edited 1 hour ago
Flux
asked 1 hour ago
FluxFlux
1145
1145
1
$begingroup$
smart programmers...
$endgroup$
– Sunnyskyguy EE75
1 hour ago
$begingroup$
There are a lot of computer architectures now and were in the past, as well as many operation systems were developed. What architecture/OS are you referring exactly? On some (old) architectures there were no possibility to stop program from what it was doing after start, that's correct. But modern hardware/OS have built in hardware tools which give only the part of the memory for the program in "normal" mode (not superuser), and it can't access memory outside this limit. Registers are free to use, as OS don't store any useful info in registers, just in memory/on disk.
$endgroup$
– cyclone125
1 hour ago
$begingroup$
In a microprocessor your program runs in "user mode", the operating systems runs in "system mode." If a user mode program executed a halt instruction, for example, the machine would not halt. The halt would be trapped and the operating system invoked. Regarding RAM, the OS would set up an environment for the user mode program so that, via memory management hardware, what the user program sees as RAM address X would not really be RAM address X.
$endgroup$
– George White
1 hour ago
$begingroup$
@cyclone125 I am not referring to any architecture/OS in particular. It's just a general question.
$endgroup$
– Flux
1 hour ago
1
$begingroup$
... Long ago I used to write in raw machine code. Yee ha !!! :-)
$endgroup$
– Russell McMahon
56 mins ago
|
show 2 more comments
1
$begingroup$
smart programmers...
$endgroup$
– Sunnyskyguy EE75
1 hour ago
$begingroup$
There are a lot of computer architectures now and were in the past, as well as many operation systems were developed. What architecture/OS are you referring exactly? On some (old) architectures there were no possibility to stop program from what it was doing after start, that's correct. But modern hardware/OS have built in hardware tools which give only the part of the memory for the program in "normal" mode (not superuser), and it can't access memory outside this limit. Registers are free to use, as OS don't store any useful info in registers, just in memory/on disk.
$endgroup$
– cyclone125
1 hour ago
$begingroup$
In a microprocessor your program runs in "user mode", the operating systems runs in "system mode." If a user mode program executed a halt instruction, for example, the machine would not halt. The halt would be trapped and the operating system invoked. Regarding RAM, the OS would set up an environment for the user mode program so that, via memory management hardware, what the user program sees as RAM address X would not really be RAM address X.
$endgroup$
– George White
1 hour ago
$begingroup$
@cyclone125 I am not referring to any architecture/OS in particular. It's just a general question.
$endgroup$
– Flux
1 hour ago
1
$begingroup$
... Long ago I used to write in raw machine code. Yee ha !!! :-)
$endgroup$
– Russell McMahon
56 mins ago
1
1
$begingroup$
smart programmers...
$endgroup$
– Sunnyskyguy EE75
1 hour ago
$begingroup$
smart programmers...
$endgroup$
– Sunnyskyguy EE75
1 hour ago
$begingroup$
There are a lot of computer architectures now and were in the past, as well as many operation systems were developed. What architecture/OS are you referring exactly? On some (old) architectures there were no possibility to stop program from what it was doing after start, that's correct. But modern hardware/OS have built in hardware tools which give only the part of the memory for the program in "normal" mode (not superuser), and it can't access memory outside this limit. Registers are free to use, as OS don't store any useful info in registers, just in memory/on disk.
$endgroup$
– cyclone125
1 hour ago
$begingroup$
There are a lot of computer architectures now and were in the past, as well as many operation systems were developed. What architecture/OS are you referring exactly? On some (old) architectures there were no possibility to stop program from what it was doing after start, that's correct. But modern hardware/OS have built in hardware tools which give only the part of the memory for the program in "normal" mode (not superuser), and it can't access memory outside this limit. Registers are free to use, as OS don't store any useful info in registers, just in memory/on disk.
$endgroup$
– cyclone125
1 hour ago
$begingroup$
In a microprocessor your program runs in "user mode", the operating systems runs in "system mode." If a user mode program executed a halt instruction, for example, the machine would not halt. The halt would be trapped and the operating system invoked. Regarding RAM, the OS would set up an environment for the user mode program so that, via memory management hardware, what the user program sees as RAM address X would not really be RAM address X.
$endgroup$
– George White
1 hour ago
$begingroup$
In a microprocessor your program runs in "user mode", the operating systems runs in "system mode." If a user mode program executed a halt instruction, for example, the machine would not halt. The halt would be trapped and the operating system invoked. Regarding RAM, the OS would set up an environment for the user mode program so that, via memory management hardware, what the user program sees as RAM address X would not really be RAM address X.
$endgroup$
– George White
1 hour ago
$begingroup$
@cyclone125 I am not referring to any architecture/OS in particular. It's just a general question.
$endgroup$
– Flux
1 hour ago
$begingroup$
@cyclone125 I am not referring to any architecture/OS in particular. It's just a general question.
$endgroup$
– Flux
1 hour ago
1
1
$begingroup$
... Long ago I used to write in raw machine code. Yee ha !!! :-)
$endgroup$
– Russell McMahon
56 mins ago
$begingroup$
... Long ago I used to write in raw machine code. Yee ha !!! :-)
$endgroup$
– Russell McMahon
56 mins ago
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
In the end, all programs are machine code, regardless of whether the source language was assembler or a high-level language.
The important thing is that there are hardware mechanisms that limit what a given process can do, including "messing with" registers that could affect other programs or the operating system itself.
This started with a simple distinction between "user" and "supervisor" modes of operation, and has since evolved into a security architecture with multiple "rings" of privilege.
Here's a very generic example to make it a little more concrete:
In "user mode", a process cannot access memory that hasn't been assigned to its process ID. Memory assigned to other processes and the operating system itself is blocked. This includes the values of registers used by those other processes. This is enforced by the MMU hardware.
Therefore, in "user mode", a process cannot access the MMU control registers.
And obviously, in "user mode", a process cannot change the mode to "supervisor mode" except through a very well-defined mechanism that involves calling an operating system function.
The operating system gets control if the process tries to break any of these rules. Once that happens, the OS can simply halt the offending process, never executing any more of its instructions.
$endgroup$
$begingroup$
If I understand correctly, what you're saying is: Some processors have "user mode" and "supervisor mode". The OS runs in "supervisor mode", and puts the processor into "user mode" to run my fictional assembly program. In "user mode", there are registers and RAM addresses that the assembly program cannot access due to deliberate design of the hardware.
$endgroup$
– Flux
1 hour ago
$begingroup$
Basically, this answer describes modern "i386-like" architectures with MMU and Protected mode. But to be true there are many old (i8080, MOS 6502 etc.) as well as modern simpler (AVR, ARM Cortex-M etc.) architectures that don't have these features and if some kind of OS is used (e.g. old CP/M, modern FreeRTOS, ChibiOS etc.) nothing can stop the program from what it is doing.
$endgroup$
– cyclone125
1 hour ago
$begingroup$
@Flux The i386 (and above) architectures provide details to learn. The x86 architecture not only has memory addresses that can be protected, but also I/O addresses, too. (There are different instructions used for memory access vs I/O access.) There are three memory addresses in the i386+. The segmented/selector-based address uses a selector register that references a table entry (GDT or LDT) to map the pair into a single 32-bit linear address. Paging tables then translate the 32-bit linear address into a 36-bit physical address (P-II.) Protection exists at both translation steps.
$endgroup$
– jonk
51 mins ago
$begingroup$
@flux your summary is correct. A program in a protected memory system with a suitable multitasking OS should not be able to crash the system with any stream of instructions. Even invalid ones - those end up at a special handler.
$endgroup$
– pjc50
27 mins ago
add a comment |
$begingroup$
Many multitasking operating systems use a data structure called a Process Control Block (PCB) to take care of the register overwrite problem. When you run your code, the OS creates a new process to keep track of it. The PCB contains information about your process and space allocated to hold register contents. Let's say process A is currently running on the processor and your code is in process B. What happens when you run your code goes something lke this:
Process A's state data (register contents, program counter, etc.) are copied into its PCB.
Process B's state data are copied from its PCB into the CPU registers
Process B runs on the processor until it finishes or is preempted
Process B's state data are copied back into its PCB
Process A's state data are copied back into the CPU and it continues running
If the operating system is running as process A, then you can see how saving its registers before your program runs then copying them back into the CPU once your program ends prevents user programs from messing with what's going in other processes.
To avoid user processes writing over OS data in memory, most platforms use memory segmentation. Basically, using virtual memory, the address space a process sees can be mapped to any arbitrary range of physical addresses. As long as the processes' physical memory spaces don't overlap, it's impossible for one process to overwrite another's data.
Of course, different OSes do things differently, so this won't apply in every case. Also, on most platforms, OS processes run in a different mode from user processes and there are some intricacies there.
New contributor
$endgroup$
add a comment |
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2 Answers
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$begingroup$
In the end, all programs are machine code, regardless of whether the source language was assembler or a high-level language.
The important thing is that there are hardware mechanisms that limit what a given process can do, including "messing with" registers that could affect other programs or the operating system itself.
This started with a simple distinction between "user" and "supervisor" modes of operation, and has since evolved into a security architecture with multiple "rings" of privilege.
Here's a very generic example to make it a little more concrete:
In "user mode", a process cannot access memory that hasn't been assigned to its process ID. Memory assigned to other processes and the operating system itself is blocked. This includes the values of registers used by those other processes. This is enforced by the MMU hardware.
Therefore, in "user mode", a process cannot access the MMU control registers.
And obviously, in "user mode", a process cannot change the mode to "supervisor mode" except through a very well-defined mechanism that involves calling an operating system function.
The operating system gets control if the process tries to break any of these rules. Once that happens, the OS can simply halt the offending process, never executing any more of its instructions.
$endgroup$
$begingroup$
If I understand correctly, what you're saying is: Some processors have "user mode" and "supervisor mode". The OS runs in "supervisor mode", and puts the processor into "user mode" to run my fictional assembly program. In "user mode", there are registers and RAM addresses that the assembly program cannot access due to deliberate design of the hardware.
$endgroup$
– Flux
1 hour ago
$begingroup$
Basically, this answer describes modern "i386-like" architectures with MMU and Protected mode. But to be true there are many old (i8080, MOS 6502 etc.) as well as modern simpler (AVR, ARM Cortex-M etc.) architectures that don't have these features and if some kind of OS is used (e.g. old CP/M, modern FreeRTOS, ChibiOS etc.) nothing can stop the program from what it is doing.
$endgroup$
– cyclone125
1 hour ago
$begingroup$
@Flux The i386 (and above) architectures provide details to learn. The x86 architecture not only has memory addresses that can be protected, but also I/O addresses, too. (There are different instructions used for memory access vs I/O access.) There are three memory addresses in the i386+. The segmented/selector-based address uses a selector register that references a table entry (GDT or LDT) to map the pair into a single 32-bit linear address. Paging tables then translate the 32-bit linear address into a 36-bit physical address (P-II.) Protection exists at both translation steps.
$endgroup$
– jonk
51 mins ago
$begingroup$
@flux your summary is correct. A program in a protected memory system with a suitable multitasking OS should not be able to crash the system with any stream of instructions. Even invalid ones - those end up at a special handler.
$endgroup$
– pjc50
27 mins ago
add a comment |
$begingroup$
In the end, all programs are machine code, regardless of whether the source language was assembler or a high-level language.
The important thing is that there are hardware mechanisms that limit what a given process can do, including "messing with" registers that could affect other programs or the operating system itself.
This started with a simple distinction between "user" and "supervisor" modes of operation, and has since evolved into a security architecture with multiple "rings" of privilege.
Here's a very generic example to make it a little more concrete:
In "user mode", a process cannot access memory that hasn't been assigned to its process ID. Memory assigned to other processes and the operating system itself is blocked. This includes the values of registers used by those other processes. This is enforced by the MMU hardware.
Therefore, in "user mode", a process cannot access the MMU control registers.
And obviously, in "user mode", a process cannot change the mode to "supervisor mode" except through a very well-defined mechanism that involves calling an operating system function.
The operating system gets control if the process tries to break any of these rules. Once that happens, the OS can simply halt the offending process, never executing any more of its instructions.
$endgroup$
$begingroup$
If I understand correctly, what you're saying is: Some processors have "user mode" and "supervisor mode". The OS runs in "supervisor mode", and puts the processor into "user mode" to run my fictional assembly program. In "user mode", there are registers and RAM addresses that the assembly program cannot access due to deliberate design of the hardware.
$endgroup$
– Flux
1 hour ago
$begingroup$
Basically, this answer describes modern "i386-like" architectures with MMU and Protected mode. But to be true there are many old (i8080, MOS 6502 etc.) as well as modern simpler (AVR, ARM Cortex-M etc.) architectures that don't have these features and if some kind of OS is used (e.g. old CP/M, modern FreeRTOS, ChibiOS etc.) nothing can stop the program from what it is doing.
$endgroup$
– cyclone125
1 hour ago
$begingroup$
@Flux The i386 (and above) architectures provide details to learn. The x86 architecture not only has memory addresses that can be protected, but also I/O addresses, too. (There are different instructions used for memory access vs I/O access.) There are three memory addresses in the i386+. The segmented/selector-based address uses a selector register that references a table entry (GDT or LDT) to map the pair into a single 32-bit linear address. Paging tables then translate the 32-bit linear address into a 36-bit physical address (P-II.) Protection exists at both translation steps.
$endgroup$
– jonk
51 mins ago
$begingroup$
@flux your summary is correct. A program in a protected memory system with a suitable multitasking OS should not be able to crash the system with any stream of instructions. Even invalid ones - those end up at a special handler.
$endgroup$
– pjc50
27 mins ago
add a comment |
$begingroup$
In the end, all programs are machine code, regardless of whether the source language was assembler or a high-level language.
The important thing is that there are hardware mechanisms that limit what a given process can do, including "messing with" registers that could affect other programs or the operating system itself.
This started with a simple distinction between "user" and "supervisor" modes of operation, and has since evolved into a security architecture with multiple "rings" of privilege.
Here's a very generic example to make it a little more concrete:
In "user mode", a process cannot access memory that hasn't been assigned to its process ID. Memory assigned to other processes and the operating system itself is blocked. This includes the values of registers used by those other processes. This is enforced by the MMU hardware.
Therefore, in "user mode", a process cannot access the MMU control registers.
And obviously, in "user mode", a process cannot change the mode to "supervisor mode" except through a very well-defined mechanism that involves calling an operating system function.
The operating system gets control if the process tries to break any of these rules. Once that happens, the OS can simply halt the offending process, never executing any more of its instructions.
$endgroup$
In the end, all programs are machine code, regardless of whether the source language was assembler or a high-level language.
The important thing is that there are hardware mechanisms that limit what a given process can do, including "messing with" registers that could affect other programs or the operating system itself.
This started with a simple distinction between "user" and "supervisor" modes of operation, and has since evolved into a security architecture with multiple "rings" of privilege.
Here's a very generic example to make it a little more concrete:
In "user mode", a process cannot access memory that hasn't been assigned to its process ID. Memory assigned to other processes and the operating system itself is blocked. This includes the values of registers used by those other processes. This is enforced by the MMU hardware.
Therefore, in "user mode", a process cannot access the MMU control registers.
And obviously, in "user mode", a process cannot change the mode to "supervisor mode" except through a very well-defined mechanism that involves calling an operating system function.
The operating system gets control if the process tries to break any of these rules. Once that happens, the OS can simply halt the offending process, never executing any more of its instructions.
edited 1 hour ago
answered 1 hour ago
Dave Tweed♦Dave Tweed
120k9149257
120k9149257
$begingroup$
If I understand correctly, what you're saying is: Some processors have "user mode" and "supervisor mode". The OS runs in "supervisor mode", and puts the processor into "user mode" to run my fictional assembly program. In "user mode", there are registers and RAM addresses that the assembly program cannot access due to deliberate design of the hardware.
$endgroup$
– Flux
1 hour ago
$begingroup$
Basically, this answer describes modern "i386-like" architectures with MMU and Protected mode. But to be true there are many old (i8080, MOS 6502 etc.) as well as modern simpler (AVR, ARM Cortex-M etc.) architectures that don't have these features and if some kind of OS is used (e.g. old CP/M, modern FreeRTOS, ChibiOS etc.) nothing can stop the program from what it is doing.
$endgroup$
– cyclone125
1 hour ago
$begingroup$
@Flux The i386 (and above) architectures provide details to learn. The x86 architecture not only has memory addresses that can be protected, but also I/O addresses, too. (There are different instructions used for memory access vs I/O access.) There are three memory addresses in the i386+. The segmented/selector-based address uses a selector register that references a table entry (GDT or LDT) to map the pair into a single 32-bit linear address. Paging tables then translate the 32-bit linear address into a 36-bit physical address (P-II.) Protection exists at both translation steps.
$endgroup$
– jonk
51 mins ago
$begingroup$
@flux your summary is correct. A program in a protected memory system with a suitable multitasking OS should not be able to crash the system with any stream of instructions. Even invalid ones - those end up at a special handler.
$endgroup$
– pjc50
27 mins ago
add a comment |
$begingroup$
If I understand correctly, what you're saying is: Some processors have "user mode" and "supervisor mode". The OS runs in "supervisor mode", and puts the processor into "user mode" to run my fictional assembly program. In "user mode", there are registers and RAM addresses that the assembly program cannot access due to deliberate design of the hardware.
$endgroup$
– Flux
1 hour ago
$begingroup$
Basically, this answer describes modern "i386-like" architectures with MMU and Protected mode. But to be true there are many old (i8080, MOS 6502 etc.) as well as modern simpler (AVR, ARM Cortex-M etc.) architectures that don't have these features and if some kind of OS is used (e.g. old CP/M, modern FreeRTOS, ChibiOS etc.) nothing can stop the program from what it is doing.
$endgroup$
– cyclone125
1 hour ago
$begingroup$
@Flux The i386 (and above) architectures provide details to learn. The x86 architecture not only has memory addresses that can be protected, but also I/O addresses, too. (There are different instructions used for memory access vs I/O access.) There are three memory addresses in the i386+. The segmented/selector-based address uses a selector register that references a table entry (GDT or LDT) to map the pair into a single 32-bit linear address. Paging tables then translate the 32-bit linear address into a 36-bit physical address (P-II.) Protection exists at both translation steps.
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– jonk
51 mins ago
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@flux your summary is correct. A program in a protected memory system with a suitable multitasking OS should not be able to crash the system with any stream of instructions. Even invalid ones - those end up at a special handler.
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– pjc50
27 mins ago
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If I understand correctly, what you're saying is: Some processors have "user mode" and "supervisor mode". The OS runs in "supervisor mode", and puts the processor into "user mode" to run my fictional assembly program. In "user mode", there are registers and RAM addresses that the assembly program cannot access due to deliberate design of the hardware.
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– Flux
1 hour ago
$begingroup$
If I understand correctly, what you're saying is: Some processors have "user mode" and "supervisor mode". The OS runs in "supervisor mode", and puts the processor into "user mode" to run my fictional assembly program. In "user mode", there are registers and RAM addresses that the assembly program cannot access due to deliberate design of the hardware.
$endgroup$
– Flux
1 hour ago
$begingroup$
Basically, this answer describes modern "i386-like" architectures with MMU and Protected mode. But to be true there are many old (i8080, MOS 6502 etc.) as well as modern simpler (AVR, ARM Cortex-M etc.) architectures that don't have these features and if some kind of OS is used (e.g. old CP/M, modern FreeRTOS, ChibiOS etc.) nothing can stop the program from what it is doing.
$endgroup$
– cyclone125
1 hour ago
$begingroup$
Basically, this answer describes modern "i386-like" architectures with MMU and Protected mode. But to be true there are many old (i8080, MOS 6502 etc.) as well as modern simpler (AVR, ARM Cortex-M etc.) architectures that don't have these features and if some kind of OS is used (e.g. old CP/M, modern FreeRTOS, ChibiOS etc.) nothing can stop the program from what it is doing.
$endgroup$
– cyclone125
1 hour ago
$begingroup$
@Flux The i386 (and above) architectures provide details to learn. The x86 architecture not only has memory addresses that can be protected, but also I/O addresses, too. (There are different instructions used for memory access vs I/O access.) There are three memory addresses in the i386+. The segmented/selector-based address uses a selector register that references a table entry (GDT or LDT) to map the pair into a single 32-bit linear address. Paging tables then translate the 32-bit linear address into a 36-bit physical address (P-II.) Protection exists at both translation steps.
$endgroup$
– jonk
51 mins ago
$begingroup$
@Flux The i386 (and above) architectures provide details to learn. The x86 architecture not only has memory addresses that can be protected, but also I/O addresses, too. (There are different instructions used for memory access vs I/O access.) There are three memory addresses in the i386+. The segmented/selector-based address uses a selector register that references a table entry (GDT or LDT) to map the pair into a single 32-bit linear address. Paging tables then translate the 32-bit linear address into a 36-bit physical address (P-II.) Protection exists at both translation steps.
$endgroup$
– jonk
51 mins ago
$begingroup$
@flux your summary is correct. A program in a protected memory system with a suitable multitasking OS should not be able to crash the system with any stream of instructions. Even invalid ones - those end up at a special handler.
$endgroup$
– pjc50
27 mins ago
$begingroup$
@flux your summary is correct. A program in a protected memory system with a suitable multitasking OS should not be able to crash the system with any stream of instructions. Even invalid ones - those end up at a special handler.
$endgroup$
– pjc50
27 mins ago
add a comment |
$begingroup$
Many multitasking operating systems use a data structure called a Process Control Block (PCB) to take care of the register overwrite problem. When you run your code, the OS creates a new process to keep track of it. The PCB contains information about your process and space allocated to hold register contents. Let's say process A is currently running on the processor and your code is in process B. What happens when you run your code goes something lke this:
Process A's state data (register contents, program counter, etc.) are copied into its PCB.
Process B's state data are copied from its PCB into the CPU registers
Process B runs on the processor until it finishes or is preempted
Process B's state data are copied back into its PCB
Process A's state data are copied back into the CPU and it continues running
If the operating system is running as process A, then you can see how saving its registers before your program runs then copying them back into the CPU once your program ends prevents user programs from messing with what's going in other processes.
To avoid user processes writing over OS data in memory, most platforms use memory segmentation. Basically, using virtual memory, the address space a process sees can be mapped to any arbitrary range of physical addresses. As long as the processes' physical memory spaces don't overlap, it's impossible for one process to overwrite another's data.
Of course, different OSes do things differently, so this won't apply in every case. Also, on most platforms, OS processes run in a different mode from user processes and there are some intricacies there.
New contributor
$endgroup$
add a comment |
$begingroup$
Many multitasking operating systems use a data structure called a Process Control Block (PCB) to take care of the register overwrite problem. When you run your code, the OS creates a new process to keep track of it. The PCB contains information about your process and space allocated to hold register contents. Let's say process A is currently running on the processor and your code is in process B. What happens when you run your code goes something lke this:
Process A's state data (register contents, program counter, etc.) are copied into its PCB.
Process B's state data are copied from its PCB into the CPU registers
Process B runs on the processor until it finishes or is preempted
Process B's state data are copied back into its PCB
Process A's state data are copied back into the CPU and it continues running
If the operating system is running as process A, then you can see how saving its registers before your program runs then copying them back into the CPU once your program ends prevents user programs from messing with what's going in other processes.
To avoid user processes writing over OS data in memory, most platforms use memory segmentation. Basically, using virtual memory, the address space a process sees can be mapped to any arbitrary range of physical addresses. As long as the processes' physical memory spaces don't overlap, it's impossible for one process to overwrite another's data.
Of course, different OSes do things differently, so this won't apply in every case. Also, on most platforms, OS processes run in a different mode from user processes and there are some intricacies there.
New contributor
$endgroup$
add a comment |
$begingroup$
Many multitasking operating systems use a data structure called a Process Control Block (PCB) to take care of the register overwrite problem. When you run your code, the OS creates a new process to keep track of it. The PCB contains information about your process and space allocated to hold register contents. Let's say process A is currently running on the processor and your code is in process B. What happens when you run your code goes something lke this:
Process A's state data (register contents, program counter, etc.) are copied into its PCB.
Process B's state data are copied from its PCB into the CPU registers
Process B runs on the processor until it finishes or is preempted
Process B's state data are copied back into its PCB
Process A's state data are copied back into the CPU and it continues running
If the operating system is running as process A, then you can see how saving its registers before your program runs then copying them back into the CPU once your program ends prevents user programs from messing with what's going in other processes.
To avoid user processes writing over OS data in memory, most platforms use memory segmentation. Basically, using virtual memory, the address space a process sees can be mapped to any arbitrary range of physical addresses. As long as the processes' physical memory spaces don't overlap, it's impossible for one process to overwrite another's data.
Of course, different OSes do things differently, so this won't apply in every case. Also, on most platforms, OS processes run in a different mode from user processes and there are some intricacies there.
New contributor
$endgroup$
Many multitasking operating systems use a data structure called a Process Control Block (PCB) to take care of the register overwrite problem. When you run your code, the OS creates a new process to keep track of it. The PCB contains information about your process and space allocated to hold register contents. Let's say process A is currently running on the processor and your code is in process B. What happens when you run your code goes something lke this:
Process A's state data (register contents, program counter, etc.) are copied into its PCB.
Process B's state data are copied from its PCB into the CPU registers
Process B runs on the processor until it finishes or is preempted
Process B's state data are copied back into its PCB
Process A's state data are copied back into the CPU and it continues running
If the operating system is running as process A, then you can see how saving its registers before your program runs then copying them back into the CPU once your program ends prevents user programs from messing with what's going in other processes.
To avoid user processes writing over OS data in memory, most platforms use memory segmentation. Basically, using virtual memory, the address space a process sees can be mapped to any arbitrary range of physical addresses. As long as the processes' physical memory spaces don't overlap, it's impossible for one process to overwrite another's data.
Of course, different OSes do things differently, so this won't apply in every case. Also, on most platforms, OS processes run in a different mode from user processes and there are some intricacies there.
New contributor
edited 43 mins ago
New contributor
answered 51 mins ago
jtstjtst
114
114
New contributor
New contributor
add a comment |
add a comment |
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$begingroup$
smart programmers...
$endgroup$
– Sunnyskyguy EE75
1 hour ago
$begingroup$
There are a lot of computer architectures now and were in the past, as well as many operation systems were developed. What architecture/OS are you referring exactly? On some (old) architectures there were no possibility to stop program from what it was doing after start, that's correct. But modern hardware/OS have built in hardware tools which give only the part of the memory for the program in "normal" mode (not superuser), and it can't access memory outside this limit. Registers are free to use, as OS don't store any useful info in registers, just in memory/on disk.
$endgroup$
– cyclone125
1 hour ago
$begingroup$
In a microprocessor your program runs in "user mode", the operating systems runs in "system mode." If a user mode program executed a halt instruction, for example, the machine would not halt. The halt would be trapped and the operating system invoked. Regarding RAM, the OS would set up an environment for the user mode program so that, via memory management hardware, what the user program sees as RAM address X would not really be RAM address X.
$endgroup$
– George White
1 hour ago
$begingroup$
@cyclone125 I am not referring to any architecture/OS in particular. It's just a general question.
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– Flux
1 hour ago
1
$begingroup$
... Long ago I used to write in raw machine code. Yee ha !!! :-)
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– Russell McMahon
56 mins ago