C Stack allocation [duplicate]

This question already has answers here: Closed 11 years ago.

Possible Duplicate:

Why does this Seg Fault?

Is the stack allocation is read only:

char* arr="abc";
arr[0]='c';

Can you change the string that is allocated on the sta开发者_StackOverflow中文版ck??

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The string "abc" isn't on the stack. A pointer to it (arr) is. Modifying the string literal is undefined behaviour.

You can see this quite clearly in the asm GCC generates on x86:

        .file   "test.c"
        .section        .rodata
.LC0:
        .string "abc"             ; String literal inside .rodata section
        .text
.globl main
        .type   main, @function
main:
        pushl   %ebp
        movl    %esp, %ebp
        subl    $16, %esp
        movl    $.LC0, -4(%ebp)   ; Pointer to LC0 (our string onto stack)
        movl    -4(%ebp), %eax    ; Pointer is copied into eax register
        movb    $99, (%eax)       ; Copy $99 ('c') to what eax points to (in .rodata)

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Your code doesn't allocate a string on the stack. It allocates a char* on the stack, that is to say a pointer, and it makes that pointer point at a string literal. Attempting to modify the string literal is undefined behavior.

To allocate the string on the stack, do:

char arr[] = "abc";

Now you've taken a copy of the string literal in your stack-allocated array arr, and you're allowed to modify that copy.

For full pedantry: everything I've described as "stack-allocated" are technically "automatic variables". C itself doesn't care where they're allocated, but I can guess with a lot of confidence that your implementation in fact does put them on a stack.

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"abc" is not allocated on the stack, it is a string literal.

No, you can't modify it. Your compiler can put that string in a read-only memory segment (if your implementation has such a concept). Trying to change it leads to undefined behavior.

(It crashes on Linux with GCC with default compile options for instance.)