Illogical bus error from printf in C

When I compile and run my code, I get a bus error right after it prints "starting." Here is what happens:

bash-3.2$ ./remDup

starting

Bus error

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

void removeDups(char* str) 
{
    int len = strlen(str);
    int i = 0;

    for (i = 0; i < len; i++) {
        char a = str[i];
        int k = i + 1;
        int c = 0;
        int j = 0;

      开发者_如何学Go  for (j = k; j < len; j++) {
            if (a != str[j]) {
                str[k] = str[j];
                k++;
            } else c++;
        }

        len -= c;
    }

    str[len] = '\0';
}

int main(int argc, const char* argv[] )
{
    char *str1 = "apple";

    printf("%s -> ", str1);
    removeDups(str1);
    printf("%s\n ", str1);

    return 1;
}

---

If you define a string as:

char *str1 = "apple";

you are not permitted to modify the contents - the standard is quite clear that this is undefined behaviour ^(a). Use:

char str1[] = "apple";

instead and it will give you a modifiable copy. It's functionally equivalent to:

char str1[6]; strcpy (str1, "apple");

---

^(a) C99 6.4.5 "String literals", paragraph 6 states:

It is unspecified whether these arrays are distinct provided their elements have the appropriate values. If the program attempts to modify such an array, the behavior is undefined.

---

You're modifying string literals which often reside in read-only memory. The standard also states that attempting to modify literals is undefined behavior.

When you're using pointers to string literals, you should either declare them as const, const char * str="text"; or as arrays char str[] = "text";

Change to e.g:

char str1[] = "apple";

In this case the compiler will create an array on stack, and copy the read-only string literal into it.