在C中修改const char *

Question

I am practicing for interviews. The problem Im currently stuck on is reversing a constant string in C. I know that since str2 is const, I can modify the location str2 points to, but not its value. I have a function below called reverse_const. It will reverse the const char *str_const in place and print it. However, when I try to print st2 after reversal from the main method, the string isn't reversed anymore. Its like reverse_const() is temporarily changing the memory location of str2. What am I doing wrong here?

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

void reverse(char *str){
    int c_size = strlen(str);
    char *c_begin = str, *c_end = str + (c_size - 1);

    int i;
    for(i = 0; i < c_size / 2; i++){
        *c_begin ^= *c_end;
        *c_end ^= *c_begin;
        *c_begin ^= *c_end;

        c_begin++;
        c_end--;
    }
}

void reverse_const(const char *str_const){
    int c_size = strlen(str_const);
    char str[c_size];
    strcpy(str, str_const);
    char *c_begin = str, *c_end = str + (c_size - 1);

    int i;
    for(i = 0; i < c_size / 2; i++){
        *c_begin ^= *c_end;
        *c_end ^= *c_begin;
        *c_begin ^= *c_end;

        c_begin++;
        c_end--;
    }

    str_const = str;
    printf("%s\n", str_const);
}

int main(){
    char str1[] = "Indiana";
    char *str2 = "Kentucky";
    printf("TESTS:\nString 1 pre-reversal: %s\n", str1);
    reverse(str1);
    printf("String 1 post-reversal: %s\n", str1);
    printf("Constant string 2 pre-reversal: %s\n", str2);
    reverse_const(str2);
    printf("Constant string 2 post-reversal: %s\n", str2);
}

Answer 1

When you pass by arguments, the function gets a copy. str_const = str; assigns to that copy. You could pass a pointer to pointer to be able change the value of a pointer outside of your function, however you're allocating the string copy on the stack, and therefore the string copy becomes invalid once you leave the scope of reverse_const so this wouldn't make sense here.

If you want a string copy that survives the end of the reverse_const , allocate the array with malloc or do both the allocation and the copying with strdup . That you have to return the malloc'ed pointer somehow (by a return value of via a pointer to pointer argument) and the caller will then have the responsibility to free that malloc 'ed memory once it's done with it.

Answer 2

You need to write your function so that you have some way to return the argument modified.

One such solution is pass-by-reference :

void reverse_const(const char **str_const){
    const char *in = *str_const;
    char *out = malloc(strlen(in)+1);

    /* write to out */

    *str_const = out;
}

But a better way would be to use the return value:

char *reverse_const(const char *str_const){
    const char *in = str_const;
    char *out = malloc(strlen(in)+1);

    /* write to out */

    return out;
}

The return type could be const char * , but that's an unnecessary restriction since you know the returned string may be safely modified.

Note that both examples use malloc . Arrays cannot be passed back, in this manner, in C, as they exist on the stack and are destroyed when the function exits.

Whenever a long-running program uses malloc , there really ought to be a matching free somewhere in the code, or else you'll have a memory leak. It's a pain, but something all C programmers must master.

Answer 3

If you want the reversed str2 back in main() , you will either need to pass an adequately sized buffer to reverse_const to hold the reversed string, or you will need to dynamically allocate storage for it in reverse_const (a local variable length array won't do):

#include <stdlib.h>
...
void reverse_const (const char **str_const)
{
    int c_size = strlen (*str_const);
    char *str = calloc (c_size + 1, sizeof *str);
    strcpy (str, *str_const);
    char *c_begin = str, *c_end = str + (c_size - 1);

    int i;
    for (i = 0; i < c_size / 2; i++) {
        *c_begin ^= *c_end;
        *c_end ^= *c_begin;
        *c_begin ^= *c_end;

        c_begin++;
        c_end--;
    }

    *str_const = str;
    printf ("%s\n", *str_const);
}

int main (void) {

    char str1[] = "Indiana";
    char *str2 = "Kentucky";

    printf ("TESTS:\nString 1 pre-reversal: %s\n", str1);

    reverse (str1);

    printf ("String 1 post-reversal: %s\n", str1);
    printf ("Constant string 2 pre-reversal: %s\n", str2);

    reverse_const ((const char **)&str2);

    printf ("Constant string 2 post-reversal: %s\n", str2);

    free (str2);

    return 0;
}

Output

$ ./bin/revconststr
TESTS:
String 1 pre-reversal: Indiana
String 1 post-reversal: anaidnI
Constant string 2 pre-reversal: Kentucky
ykcutneK
Constant string 2 post-reversal: ykcutneK

Returning The Pointer

You have an additional option to return the pointer to str to assign to str2 in main() . This is more what you would normally expect to see. Let me know if you have any questions:

char *reverse_const2 (const char **str_const)
{
    int c_size = strlen (*str_const);
    char *str = calloc (c_size + 1, sizeof *str);
    strcpy (str, *str_const);
    char *c_begin = str, *c_end = str + (c_size - 1);

    int i;
    for (i = 0; i < c_size / 2; i++) {
        *c_begin ^= *c_end;
        *c_end ^= *c_begin;
        *c_begin ^= *c_end;

        c_begin++;
        c_end--;
    }

    //*str_const = str;
    printf ("%s\n", *str_const);

    return str;
}

int main (void)
{

    char str1[] = "Indiana";
    char *str2 = "Kentucky";

    printf ("TESTS:\nString 1 pre-reversal: %s\n", str1);

    reverse (str1);

    printf ("String 1 post-reversal: %s\n", str1);
    printf ("Constant string 2 pre-reversal: %s\n", str2);

    str2 = reverse_const2 ((const char **)&str2);

    printf ("Constant string 2 post-reversal: %s\n", str2);

    free (str2);

    return 0;
}

Answer 4

The problem here is that you modify the argument passed to reverse_const , but arguments in C are passed by value meaning that they are copied. The variable you modify in the function is a copy of the original pointer, changing a copy will of course not change the original.

C doesn't have pass by reference which is needed here, but it can be emulated by using pointers, in the case of your function you need to pass a pointer to the pointer, then use the dereference operator * to modify what the pointer to pointer points to, and use the address-of operator & when calling the function.