Xcode Exc_BAD_ACCESS

Question

so this code is giving me the Exc_bad_access_code(2) error and i have no idea why. I think the problem is with the parameters but im not sure, any thoughts?

#include <stdio.h>

void swap(char *x, char *y);


/* Function to swap values at two pointers */
void swap(char *x, char *y)
{
    char temp;
    temp = *x;
    *x = *y;
    *y = temp;
}

int main(int argc, const char * argv[]) {
    // insert code here...

    char *a = "ASD123";
    swap (a+1 , a+2);

    return 0;
}

Answer 1

You need to post the complete crash log to be sure, but it's probably that you are attempting to manipulate a string constant:

char *a = "ASD123";
swap (a+1 , a+2);

which probably lives in read-only memory.

Try:

char a[12];
strcpy(a, "ASD123");
swap (a+1 , a+2);

or:

char a[] = "ASD123";
swap (a+1 , a+2);

That will copy the string onto the stack, where it may be modified without issue. You could also use strdup() to copy the string onto the heap (don't forget to call free() to release the allocated memory).

Answer 2

You need to initialize a correctly. Something like char a[7] = "ASD123";

char* a only creates a pointer-to-char , it does not allocate enough memory to store the characters.

Answer 3

You are trying to modify a string literal.

Change

char *a = "ASD123";

to

char a[] = "ASD123";

and your program will work.

In the first case a is simply a pointer pointing to a string literal which is in a write protected part of the memory on most platforms, hence the crash because you try to write info write protected memory.

In the second case a is a char array of length 7 which is initialized with "ASD123" (6 chars for the string and one char for the terminating zero).