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strToHex ( string to its hex representation as string)


Integer-to-hex string generatorConvert hex string to byte arrayString representation of TreeSimple string inverter programSimple string repeater programGeneric Pairing Heap PerformanceHex-string representation of a byte arrayConvert hex color string to SDL ColorC binary number converter (Hex and Decimals)Convert string of hex into vector of bytes






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}







1












$begingroup$


I want to convert strings to their hex representations as strings too (like hex dump programs), for example "abz" to "61627A".



char * strToHex( char * str )
{
int length = strlen ( str );
char * newStr = malloc( length * 2 );
if ( !newStr ) shutDown ( "can't alloc memory" ) ;

for ( int x = 0; x < length; x++){
char y = str[ x ];
sprintf ( newStr + x * 2, "%02X", y );
}
return newStr;
}


ShutDown definition is omitted here, it is a function that calls perror and exit()



I designed strToHex to be used like



char * str = "abcdefghijklmnopqrstuvwxyz";
char * hex = strToHex(str);
printf("%sn",hex);
//outputs : 6162636465666768696A6B6C6D6E6F707172737475767778797A









share|improve this question











$endgroup$








  • 2




    $begingroup$
    I'd be really interested to see what shutdown(char* msg) does.
    $endgroup$
    – pacmaninbw
    1 hour ago










  • $begingroup$
    In the use case that was provided, since you can effectively predict the size, I would think it would be more natural to have a string buffer and the size passed in instead of creating it dynamically.
    $endgroup$
    – Neil Edelman
    39 mins ago


















1












$begingroup$


I want to convert strings to their hex representations as strings too (like hex dump programs), for example "abz" to "61627A".



char * strToHex( char * str )
{
int length = strlen ( str );
char * newStr = malloc( length * 2 );
if ( !newStr ) shutDown ( "can't alloc memory" ) ;

for ( int x = 0; x < length; x++){
char y = str[ x ];
sprintf ( newStr + x * 2, "%02X", y );
}
return newStr;
}


ShutDown definition is omitted here, it is a function that calls perror and exit()



I designed strToHex to be used like



char * str = "abcdefghijklmnopqrstuvwxyz";
char * hex = strToHex(str);
printf("%sn",hex);
//outputs : 6162636465666768696A6B6C6D6E6F707172737475767778797A









share|improve this question











$endgroup$








  • 2




    $begingroup$
    I'd be really interested to see what shutdown(char* msg) does.
    $endgroup$
    – pacmaninbw
    1 hour ago










  • $begingroup$
    In the use case that was provided, since you can effectively predict the size, I would think it would be more natural to have a string buffer and the size passed in instead of creating it dynamically.
    $endgroup$
    – Neil Edelman
    39 mins ago














1












1








1





$begingroup$


I want to convert strings to their hex representations as strings too (like hex dump programs), for example "abz" to "61627A".



char * strToHex( char * str )
{
int length = strlen ( str );
char * newStr = malloc( length * 2 );
if ( !newStr ) shutDown ( "can't alloc memory" ) ;

for ( int x = 0; x < length; x++){
char y = str[ x ];
sprintf ( newStr + x * 2, "%02X", y );
}
return newStr;
}


ShutDown definition is omitted here, it is a function that calls perror and exit()



I designed strToHex to be used like



char * str = "abcdefghijklmnopqrstuvwxyz";
char * hex = strToHex(str);
printf("%sn",hex);
//outputs : 6162636465666768696A6B6C6D6E6F707172737475767778797A









share|improve this question











$endgroup$




I want to convert strings to their hex representations as strings too (like hex dump programs), for example "abz" to "61627A".



char * strToHex( char * str )
{
int length = strlen ( str );
char * newStr = malloc( length * 2 );
if ( !newStr ) shutDown ( "can't alloc memory" ) ;

for ( int x = 0; x < length; x++){
char y = str[ x ];
sprintf ( newStr + x * 2, "%02X", y );
}
return newStr;
}


ShutDown definition is omitted here, it is a function that calls perror and exit()



I designed strToHex to be used like



char * str = "abcdefghijklmnopqrstuvwxyz";
char * hex = strToHex(str);
printf("%sn",hex);
//outputs : 6162636465666768696A6B6C6D6E6F707172737475767778797A






beginner c strings






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 min ago









mdfst13

17.9k62257




17.9k62257










asked 3 hours ago









Accountant مAccountant م

1827




1827








  • 2




    $begingroup$
    I'd be really interested to see what shutdown(char* msg) does.
    $endgroup$
    – pacmaninbw
    1 hour ago










  • $begingroup$
    In the use case that was provided, since you can effectively predict the size, I would think it would be more natural to have a string buffer and the size passed in instead of creating it dynamically.
    $endgroup$
    – Neil Edelman
    39 mins ago














  • 2




    $begingroup$
    I'd be really interested to see what shutdown(char* msg) does.
    $endgroup$
    – pacmaninbw
    1 hour ago










  • $begingroup$
    In the use case that was provided, since you can effectively predict the size, I would think it would be more natural to have a string buffer and the size passed in instead of creating it dynamically.
    $endgroup$
    – Neil Edelman
    39 mins ago








2




2




$begingroup$
I'd be really interested to see what shutdown(char* msg) does.
$endgroup$
– pacmaninbw
1 hour ago




$begingroup$
I'd be really interested to see what shutdown(char* msg) does.
$endgroup$
– pacmaninbw
1 hour ago












$begingroup$
In the use case that was provided, since you can effectively predict the size, I would think it would be more natural to have a string buffer and the size passed in instead of creating it dynamically.
$endgroup$
– Neil Edelman
39 mins ago




$begingroup$
In the use case that was provided, since you can effectively predict the size, I would think it would be more natural to have a string buffer and the size passed in instead of creating it dynamically.
$endgroup$
– Neil Edelman
39 mins ago










1 Answer
1






active

oldest

votes


















2












$begingroup$

Formatting



Most C formatting guides do not include spaces around the arguments to function calls, nor the expressions within an if-statement. For an example of a C style most C programmers would find acceptable, see OpenBSD's style(9) manual.



I choose to associate * with the variable name, rather than floating between the type and name. This disambiguates the following example:



int *a, b;


Here, a is a pointer to an integer, but b is only an integer. By moving the asterisk next to the name, it makes this clearer.



int length = strlen ( str );
char * newStr = malloc (length * 2 );
if ( !newStr) shutDown ( "can't allocate memory" ) ;


Becomes:



int const len = strlen(str);
char *const new_str = malloc(len * 2);

if (new_str == NULL) {
shutDown("can't allocate memory");
}


Error checking



Rather than calling shutDown() and exit()ing the program, you should instead return an error value which can be checked by the caller of str_to_hex(). Because you return a pointer, you can return NULL to indicate an error occurred and the caller should check errno.



Likewise, on some systems your program can incorrectly exit when length == 0. If we look at the manual page for malloc(3):




Return Value



The malloc() and calloc() functions return a pointer to the allocated memory that is suitably aligned for any kind of variable. On error, these functions return NULL. NULL may also be returned by a successful call to malloc() with a size of zero, or by a successful call to calloc() with nmemb or size equal to zero.




So by returning NULL we account for the case where malloc(3) returns NULL on success.



if (new_str == NULL) {
shutDown("can't alloc memory");
}


Becomes:



if (new_str == NULL) {
return NULL;
}


If you choose, you can also check if str is NULL before calling strlen(). This is up to you, and it's not uncommon in C to ignore this case and leave it as user error.



Looping



Use the size_t type in your loop rather than int. size_t is guaranteed be wide enough to hold any array index, while int is not.



Using i rather than x is more common for looping variables.



The y variable isn't needed. You can simply use str[i] in its place.



In terms of performance there's likely a faster option than using sprintf(). You should look into strtol(3).



Conclusion



Here is the code I ended up with:



#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char *
str_to_hex(char const *const str)
{
size_t const len = strlen(str);

char *const new_str = malloc(len * 2);

if (new_str == NULL) {
return NULL;
}

for (size_t i = 0; i < len; ++i) {
sprintf(new_str + i * 2, "%02X", str[i]);
}

return new_str;
}

int
main(void)
{
char *str = "abz";
char *hex = str_to_hex(str);

if (hex == NULL && strlen(str) != 0) {
/* error ... */
}

printf("%sn",hex);

free(hex);
}


Hope this helps!






share|improve this answer











$endgroup$














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    1 Answer
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    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Formatting



    Most C formatting guides do not include spaces around the arguments to function calls, nor the expressions within an if-statement. For an example of a C style most C programmers would find acceptable, see OpenBSD's style(9) manual.



    I choose to associate * with the variable name, rather than floating between the type and name. This disambiguates the following example:



    int *a, b;


    Here, a is a pointer to an integer, but b is only an integer. By moving the asterisk next to the name, it makes this clearer.



    int length = strlen ( str );
    char * newStr = malloc (length * 2 );
    if ( !newStr) shutDown ( "can't allocate memory" ) ;


    Becomes:



    int const len = strlen(str);
    char *const new_str = malloc(len * 2);

    if (new_str == NULL) {
    shutDown("can't allocate memory");
    }


    Error checking



    Rather than calling shutDown() and exit()ing the program, you should instead return an error value which can be checked by the caller of str_to_hex(). Because you return a pointer, you can return NULL to indicate an error occurred and the caller should check errno.



    Likewise, on some systems your program can incorrectly exit when length == 0. If we look at the manual page for malloc(3):




    Return Value



    The malloc() and calloc() functions return a pointer to the allocated memory that is suitably aligned for any kind of variable. On error, these functions return NULL. NULL may also be returned by a successful call to malloc() with a size of zero, or by a successful call to calloc() with nmemb or size equal to zero.




    So by returning NULL we account for the case where malloc(3) returns NULL on success.



    if (new_str == NULL) {
    shutDown("can't alloc memory");
    }


    Becomes:



    if (new_str == NULL) {
    return NULL;
    }


    If you choose, you can also check if str is NULL before calling strlen(). This is up to you, and it's not uncommon in C to ignore this case and leave it as user error.



    Looping



    Use the size_t type in your loop rather than int. size_t is guaranteed be wide enough to hold any array index, while int is not.



    Using i rather than x is more common for looping variables.



    The y variable isn't needed. You can simply use str[i] in its place.



    In terms of performance there's likely a faster option than using sprintf(). You should look into strtol(3).



    Conclusion



    Here is the code I ended up with:



    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>

    char *
    str_to_hex(char const *const str)
    {
    size_t const len = strlen(str);

    char *const new_str = malloc(len * 2);

    if (new_str == NULL) {
    return NULL;
    }

    for (size_t i = 0; i < len; ++i) {
    sprintf(new_str + i * 2, "%02X", str[i]);
    }

    return new_str;
    }

    int
    main(void)
    {
    char *str = "abz";
    char *hex = str_to_hex(str);

    if (hex == NULL && strlen(str) != 0) {
    /* error ... */
    }

    printf("%sn",hex);

    free(hex);
    }


    Hope this helps!






    share|improve this answer











    $endgroup$


















      2












      $begingroup$

      Formatting



      Most C formatting guides do not include spaces around the arguments to function calls, nor the expressions within an if-statement. For an example of a C style most C programmers would find acceptable, see OpenBSD's style(9) manual.



      I choose to associate * with the variable name, rather than floating between the type and name. This disambiguates the following example:



      int *a, b;


      Here, a is a pointer to an integer, but b is only an integer. By moving the asterisk next to the name, it makes this clearer.



      int length = strlen ( str );
      char * newStr = malloc (length * 2 );
      if ( !newStr) shutDown ( "can't allocate memory" ) ;


      Becomes:



      int const len = strlen(str);
      char *const new_str = malloc(len * 2);

      if (new_str == NULL) {
      shutDown("can't allocate memory");
      }


      Error checking



      Rather than calling shutDown() and exit()ing the program, you should instead return an error value which can be checked by the caller of str_to_hex(). Because you return a pointer, you can return NULL to indicate an error occurred and the caller should check errno.



      Likewise, on some systems your program can incorrectly exit when length == 0. If we look at the manual page for malloc(3):




      Return Value



      The malloc() and calloc() functions return a pointer to the allocated memory that is suitably aligned for any kind of variable. On error, these functions return NULL. NULL may also be returned by a successful call to malloc() with a size of zero, or by a successful call to calloc() with nmemb or size equal to zero.




      So by returning NULL we account for the case where malloc(3) returns NULL on success.



      if (new_str == NULL) {
      shutDown("can't alloc memory");
      }


      Becomes:



      if (new_str == NULL) {
      return NULL;
      }


      If you choose, you can also check if str is NULL before calling strlen(). This is up to you, and it's not uncommon in C to ignore this case and leave it as user error.



      Looping



      Use the size_t type in your loop rather than int. size_t is guaranteed be wide enough to hold any array index, while int is not.



      Using i rather than x is more common for looping variables.



      The y variable isn't needed. You can simply use str[i] in its place.



      In terms of performance there's likely a faster option than using sprintf(). You should look into strtol(3).



      Conclusion



      Here is the code I ended up with:



      #include <stdio.h>
      #include <stdlib.h>
      #include <string.h>

      char *
      str_to_hex(char const *const str)
      {
      size_t const len = strlen(str);

      char *const new_str = malloc(len * 2);

      if (new_str == NULL) {
      return NULL;
      }

      for (size_t i = 0; i < len; ++i) {
      sprintf(new_str + i * 2, "%02X", str[i]);
      }

      return new_str;
      }

      int
      main(void)
      {
      char *str = "abz";
      char *hex = str_to_hex(str);

      if (hex == NULL && strlen(str) != 0) {
      /* error ... */
      }

      printf("%sn",hex);

      free(hex);
      }


      Hope this helps!






      share|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Formatting



        Most C formatting guides do not include spaces around the arguments to function calls, nor the expressions within an if-statement. For an example of a C style most C programmers would find acceptable, see OpenBSD's style(9) manual.



        I choose to associate * with the variable name, rather than floating between the type and name. This disambiguates the following example:



        int *a, b;


        Here, a is a pointer to an integer, but b is only an integer. By moving the asterisk next to the name, it makes this clearer.



        int length = strlen ( str );
        char * newStr = malloc (length * 2 );
        if ( !newStr) shutDown ( "can't allocate memory" ) ;


        Becomes:



        int const len = strlen(str);
        char *const new_str = malloc(len * 2);

        if (new_str == NULL) {
        shutDown("can't allocate memory");
        }


        Error checking



        Rather than calling shutDown() and exit()ing the program, you should instead return an error value which can be checked by the caller of str_to_hex(). Because you return a pointer, you can return NULL to indicate an error occurred and the caller should check errno.



        Likewise, on some systems your program can incorrectly exit when length == 0. If we look at the manual page for malloc(3):




        Return Value



        The malloc() and calloc() functions return a pointer to the allocated memory that is suitably aligned for any kind of variable. On error, these functions return NULL. NULL may also be returned by a successful call to malloc() with a size of zero, or by a successful call to calloc() with nmemb or size equal to zero.




        So by returning NULL we account for the case where malloc(3) returns NULL on success.



        if (new_str == NULL) {
        shutDown("can't alloc memory");
        }


        Becomes:



        if (new_str == NULL) {
        return NULL;
        }


        If you choose, you can also check if str is NULL before calling strlen(). This is up to you, and it's not uncommon in C to ignore this case and leave it as user error.



        Looping



        Use the size_t type in your loop rather than int. size_t is guaranteed be wide enough to hold any array index, while int is not.



        Using i rather than x is more common for looping variables.



        The y variable isn't needed. You can simply use str[i] in its place.



        In terms of performance there's likely a faster option than using sprintf(). You should look into strtol(3).



        Conclusion



        Here is the code I ended up with:



        #include <stdio.h>
        #include <stdlib.h>
        #include <string.h>

        char *
        str_to_hex(char const *const str)
        {
        size_t const len = strlen(str);

        char *const new_str = malloc(len * 2);

        if (new_str == NULL) {
        return NULL;
        }

        for (size_t i = 0; i < len; ++i) {
        sprintf(new_str + i * 2, "%02X", str[i]);
        }

        return new_str;
        }

        int
        main(void)
        {
        char *str = "abz";
        char *hex = str_to_hex(str);

        if (hex == NULL && strlen(str) != 0) {
        /* error ... */
        }

        printf("%sn",hex);

        free(hex);
        }


        Hope this helps!






        share|improve this answer











        $endgroup$



        Formatting



        Most C formatting guides do not include spaces around the arguments to function calls, nor the expressions within an if-statement. For an example of a C style most C programmers would find acceptable, see OpenBSD's style(9) manual.



        I choose to associate * with the variable name, rather than floating between the type and name. This disambiguates the following example:



        int *a, b;


        Here, a is a pointer to an integer, but b is only an integer. By moving the asterisk next to the name, it makes this clearer.



        int length = strlen ( str );
        char * newStr = malloc (length * 2 );
        if ( !newStr) shutDown ( "can't allocate memory" ) ;


        Becomes:



        int const len = strlen(str);
        char *const new_str = malloc(len * 2);

        if (new_str == NULL) {
        shutDown("can't allocate memory");
        }


        Error checking



        Rather than calling shutDown() and exit()ing the program, you should instead return an error value which can be checked by the caller of str_to_hex(). Because you return a pointer, you can return NULL to indicate an error occurred and the caller should check errno.



        Likewise, on some systems your program can incorrectly exit when length == 0. If we look at the manual page for malloc(3):




        Return Value



        The malloc() and calloc() functions return a pointer to the allocated memory that is suitably aligned for any kind of variable. On error, these functions return NULL. NULL may also be returned by a successful call to malloc() with a size of zero, or by a successful call to calloc() with nmemb or size equal to zero.




        So by returning NULL we account for the case where malloc(3) returns NULL on success.



        if (new_str == NULL) {
        shutDown("can't alloc memory");
        }


        Becomes:



        if (new_str == NULL) {
        return NULL;
        }


        If you choose, you can also check if str is NULL before calling strlen(). This is up to you, and it's not uncommon in C to ignore this case and leave it as user error.



        Looping



        Use the size_t type in your loop rather than int. size_t is guaranteed be wide enough to hold any array index, while int is not.



        Using i rather than x is more common for looping variables.



        The y variable isn't needed. You can simply use str[i] in its place.



        In terms of performance there's likely a faster option than using sprintf(). You should look into strtol(3).



        Conclusion



        Here is the code I ended up with:



        #include <stdio.h>
        #include <stdlib.h>
        #include <string.h>

        char *
        str_to_hex(char const *const str)
        {
        size_t const len = strlen(str);

        char *const new_str = malloc(len * 2);

        if (new_str == NULL) {
        return NULL;
        }

        for (size_t i = 0; i < len; ++i) {
        sprintf(new_str + i * 2, "%02X", str[i]);
        }

        return new_str;
        }

        int
        main(void)
        {
        char *str = "abz";
        char *hex = str_to_hex(str);

        if (hex == NULL && strlen(str) != 0) {
        /* error ... */
        }

        printf("%sn",hex);

        free(hex);
        }


        Hope this helps!







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 31 mins ago

























        answered 46 mins ago









        esoteesote

        2,93111038




        2,93111038






























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