为什么堆栈指针向下移动比臂Linux的gnueabi-GCC编译时堆栈帧尺寸更大的4个字节？ [英] Why is the stack pointer moved down 4 bytes greater than the stack frame size when compiling with arm-linux-gnueabi-gcc?

问题描述

使用下面的琐碎的C程序作为一个例子。主（）使函数调用来概括传递4的整数。和（）使用4本地人。

Using the trivial C program below as an example. main() makes a function call to sum passing in 4 integer. sum() uses 4 locals.

void sum(int a, int b, int c, int d);

void main(void)
{
    sum(11, 12, 13, 14);
}

void sum(int a, int b, int c, int d)
{
    int x;
    int y;
    int z;
    int z2;

    x = a;
    y = b;
    z = c;
    z2 = d;
}

在我的Ubuntu服务器04年4月12日LTS我使用编译这个程序

On my Ubuntu server 12.04.04 LTS I compile this program using

ARM-Linux的gnueabi-GCC -S -mthumb func.c

arm-linux-gnueabi-gcc -S -mthumb func.c

sum:
@ args = 0, pretend = 0, frame = 32
@ frame_needed = 1, uses_anonymous_args = 0
@ link register save eliminated.
push    {r7}
sub sp, sp, #36    <===   why is this 36 and not 32 bytes?
add r7, sp, #0

str r0, [r7, #12]
str r1, [r7, #8]
str r2, [r7, #4]
str r3, [r7, #0]   <- paramaters passed

ldr r3, [r7, #12]
str r3, [r7, #16]  <- locals
ldr r3, [r7, #8]
str r3, [r7, #20]
ldr r3, [r7, #4]
str r3, [r7, #24]
ldr r3, [r7, #0]
str r3, [r7, #28]

add r7, r7, #36
mov sp, r7
pop {r7}
bx  lr

看来，INT是一个4字节的每个。 4当地人和4个参数的功能，使共（4 * 4字节）+（4 * 4字节）= 32字节，这汇编输出框= 32。匹配

It appears that int's a 4 bytes each. 4 locals and 4 arguments for the function makes a total of (4 *4 bytes) + (4 * 4bytes) = 32 bytes and this matches the assembly output "frame = 32".

但为什么堆栈指针得到由36递减，并且不只是32？

But why does the stack pointer get decremented by 36 and not just 32?

推荐答案

有关ARM过程调用标准需要8字节对齐。

Procedure call standard for ARM requires 8 byte alignment.

在一个公共接口5.2.1.2堆栈限制

5.2.1.2 Stack constraints at a public interface

该协议栈还必须符合以下约束在一个公共接口：

The stack must also conform to the following constraint at a public interface:

  
• SPmod8 = 0的纸张都必须是双字对齐。

  

既然你是生产装配一切是默认输出，让您得到8字节对齐（我想这和gcc不添加。全球＆LT;符号＆GT; 指令生成集时，我想这说，即使一个静态函数是一个的公共接口的或GCC只对准每一个功能有8字节堆栈对齐静态函数。）

Since you are producing assembly everything is exported by default, so you get 8 byte alignment. (I tried this and gcc doesn't add .global <symbol> directive to static functions when generating assembly. I guess this says even a static function is a public interface or gcc just aligns every function to have 8-byte stack alignment.)

<一个href=\"http://gcc.godbolt.org/#%7B%22version%22:3,%22filterAsm%22%3a%7B%22labels%22%3atrue,%22directives%22%3atrue,%22commentOnly%22%3atrue%7D,%22compilers%22%3a%5B%7B%22sourcez%22%3a%22G4ewlgJgBAzgrgWwBRgHYBcoEMA0U2YBGeBUAxiRlBAJQDcAUA6QlmkqJDQwN4NQDYiJAEYReEQCYJAZgkAWegwC+TTtHjJSufFWK7MFA9W59BxgB6NzpAJ7XBpAF4OBzyY36CLUALzZXKFs/KEJApxCycMkQiEZlIA=%22,%22compiler%22%3a%22/usr/bin/arm-linux-gnueabi-g++-4.6%22,%22options%22%3a%22-fomit-frame-pointer%20-mthumb%22%7D%5D%7D\"相对=nofollow>您可以使用 -fomit-frame-pointer的跳过推 R7 然后海湾合作委员会应在32离开堆栈的深度。

You can use -fomit-frame-pointer to skip pushing r7 then gcc should leave the stack depth at 32.

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