C 堆栈分配

发布于 2024-12-08 15:55:33 字数 261 浏览 0 评论 0原文

可能的重复：
为什么会出现此段错误？

堆栈分配是否为只读：

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

您能否更改在堆栈上分配的字符串？

原文

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 stack??

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翻了热茶 2024-12-15 15:55:33

字符串 "abc" 不在堆栈中。指向它的指针 (arr) 是。修改字符串文字是未定义的行为。

您可以在 GCC 在 x86 上生成的 asm 中清楚地看到这一点：

        .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)

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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孤凫 2024-12-15 15:55:33

您的代码不会在堆栈上分配字符串。它在堆栈上分配一个 char*，即一个指针，并使该指针指向字符串文字。尝试修改字符串文字是未定义的行为。

要在堆栈上分配字符串，请执行以下操作：

char arr[] = "abc";

现在，您已在堆栈分配的数组 arr 中获取了字符串文字的副本，并且您可以修改该副本。

对于完整的迂腐：我所描述的“堆栈分配”的所有内容在技术上都是“自动变量”。 C 本身并不关心它们被分配在哪里，但我可以很有信心地猜测您的实现实际上确实将它们放在了堆栈上。

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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