Why is assigning a character in a string to itself a bus error?

Question

This works and produces bbcd as I'd expect.

#include <stdio.h>

int main(void) {
    char string[] = "abcd";

    string[0] = string[1];

    printf("%s\n", string);
}

This is a bus error.

#include <stdio.h>

int main(void) {
    char *string = "abcd";

    string[0] = string[1];

    printf("%s\n", string);
}

Why?

Valgrind says:

==9909== Process terminating with default action of signal 10 (SIGBUS)
==9909==  Non-existent physical address at address 0x100000FA2
==9909==    at 0x100000F65: main (test.c:6)

`abcd` is a constant. You can do `char string[5] = "abcd";`instead. — Support Ukraine, Nov 04 '16 at 07:47
`char string[] = "abcd";` declares an array and initialises it with `{ 'a', 'b', 'c', 'd', '\0' }`. `char *string = "abcd";` declares a pointer to a `char` and initialises it with a pointer to the beginning of the **constant string** `"abcd"`. You can change the array you declared, but modifying a constant string leads to undefined behaviour. — , Nov 04 '16 at 07:49
@2501 ["char bus error"](http://stackoverflow.com/search?q=char+bus+error) turns up a lot of unrelated things. Thanks for pointing out the dup. — Schwern, Nov 04 '16 at 07:50
@PaulRooney I tried `cc -Weverything -pedantic -std=c99 test.c` and no warnings. — Schwern, Nov 04 '16 at 07:52
@Schwern Add the flag `-Wwrite-strings` (if clang supports it). — 2501, Nov 04 '16 at 07:54
@2501 Of course! `-Wall` isn't all so why would `-Weverything` be everything? \*sigh\* It didn't catch it. I got a warning about my lack of `const` which might have at least been a hint. "*initializing 'char *' with an expression of type 'const char [5]' discards qualifiers [-Wincompatible-pointer-types-discards-qualifiers]*" (I only sat down and properly learned C last winter) — Schwern, Nov 04 '16 at 07:57
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/127353/discussion-between-schwern-and-2501). — Schwern, Nov 04 '16 at 08:02

score 0 · Answer 1 · answered Nov 04 '16 at 07:47

Because in the second case, you're trying to modify a string literal which invokes undefined behavior.

To elaborate, in the second case, string[0] is the first element of the string literal, and any assignment to that is an attempt to modify the value held by that element.

Quoting C11, chapter §6.4.5, "String literals"

[...] If the program attempts to modify such an array, the behavior is undefined.

Why is assigning a character in a string to itself a bus error?

1 Answers1