关于函数和栈调用的疑惑

发布于 2023-08-12 23:53:26 浏览：296 订阅该版

1. 编译器为 arm-none-eabi-gcc  v10.3.1，优化等级为 -O0
2. 示例代码如下

```c
rt_thread_t s_test_thread = NULL;

void test_func(void)
{
    char array[10] = {0};
    array[1] = 1;
    array[9] = 9;

uint32_t num = 10;

char buf[32] = {0};

buf[15] = 15;
    buf[31] = 31;
}

void test_func2(void)
{
    char array[10] = {0};
}

void test_func3(void)
{
    char array[200] = {0};
}

void test_thread(void *args)
{
    test_func();
    test_func2();
    test_func3();
    while (1) {
        sleep(1);
    }
}

int main(void)
{
    s_test_thread = rt_thread_create("test",
                                     test_thread,
                                     NULL,
                                     4 * 1024,
                                     15,
                                     10);
    rt_thread_startup(s_test_thread);
    while (1) {
        sleep(1);
    }
    return 0;
}
```
3. 反汇编代码如下

```c
void test_func(void)
{
 8000b80:	b480      	push	{r7}
 8000b82:	b08d      	sub	sp, #52	; 0x34
 8000b84:	af00      	add	r7, sp, #0
    char array[10] = {0};
 8000b86:	2300      	movs	r3, #0
 8000b88:	623b      	str	r3, [r7, #32]
 8000b8a:	f107 0324 	add.w	r3, r7, #36	; 0x24
 8000b8e:	2200      	movs	r2, #0
 8000b90:	601a      	str	r2, [r3, #0]
 8000b92:	809a      	strh	r2, [r3, #4]
    array[1] = 1;
 8000b94:	2301      	movs	r3, #1
 8000b96:	f887 3021 	strb.w	r3, [r7, #33]	; 0x21
    array[9] = 9;
 8000b9a:	2309      	movs	r3, #9
 8000b9c:	f887 3029 	strb.w	r3, [r7, #41]	; 0x29

uint32_t num = 10;
 8000ba0:	230a      	movs	r3, #10
 8000ba2:	62fb      	str	r3, [r7, #44]	; 0x2c

char buf[32] = {0};
 8000ba4:	2300      	movs	r3, #0
 8000ba6:	603b      	str	r3, [r7, #0]
 8000ba8:	1d3b      	adds	r3, r7, #4
 8000baa:	2200      	movs	r2, #0
 8000bac:	601a      	str	r2, [r3, #0]
 8000bae:	605a      	str	r2, [r3, #4]
 8000bb0:	609a      	str	r2, [r3, #8]
 8000bb2:	60da      	str	r2, [r3, #12]
 8000bb4:	611a      	str	r2, [r3, #16]
 8000bb6:	615a      	str	r2, [r3, #20]
 8000bb8:	619a      	str	r2, [r3, #24]

buf[15] = 15;
 8000bba:	230f      	movs	r3, #15
 8000bbc:	73fb      	strb	r3, [r7, #15]
    buf[31] = 31;
 8000bbe:	231f      	movs	r3, #31
 8000bc0:	77fb      	strb	r3, [r7, #31]
}
 8000bc2:	bf00      	nop
 8000bc4:	3734      	adds	r7, #52	; 0x34
 8000bc6:	46bd      	mov	sp, r7
 8000bc8:	bc80      	pop	{r7}
 8000bca:	4770      	bx	lr

void test_func2(void)
{
 8000bcc:	b480      	push	{r7}
 8000bce:	b085      	sub	sp, #20
 8000bd0:	af00      	add	r7, sp, #0
    char array[10] = {0};
 8000bd2:	2300      	movs	r3, #0
 8000bd4:	607b      	str	r3, [r7, #4]
 8000bd6:	f107 0308 	add.w	r3, r7, #8
 8000bda:	2200      	movs	r2, #0
 8000bdc:	601a      	str	r2, [r3, #0]
 8000bde:	809a      	strh	r2, [r3, #4]
}
 8000be0:	bf00      	nop
 8000be2:	3714      	adds	r7, #20
 8000be4:	46bd      	mov	sp, r7
 8000be6:	bc80      	pop	{r7}
 8000be8:	4770      	bx	lr

void test_func3(void)
{
 8000bea:	b580      	push	{r7, lr}
 8000bec:	b0b2      	sub	sp, #200	; 0xc8
 8000bee:	af00      	add	r7, sp, #0
    char array[200] = {0};
 8000bf0:	2300      	movs	r3, #0
 8000bf2:	603b      	str	r3, [r7, #0]
 8000bf4:	1d3b      	adds	r3, r7, #4
 8000bf6:	22c4      	movs	r2, #196	; 0xc4
 8000bf8:	2100      	movs	r1, #0
 8000bfa:	4618      	mov	r0, r3
 8000bfc:	f03b fd82 	bl	803c704 <memset>
}
 8000c00:	bf00      	nop
 8000c02:	37c8      	adds	r7, #200	; 0xc8
 8000c04:	46bd      	mov	sp, r7
 8000c06:	bd80      	pop	{r7, pc}
```

**疑问**：
1. 在 `test_func` 函数中，用到的栈空间为 46（10 + 4 + 32）字节，却预留 52 字节的空间，这多出来的 6 字节的用途是什么？
2. 在 `test_func2` 函数中，用到的栈空间为 10 字节，却预留 20 字节的空间，这多出来的 10 字节的用途是什么？
3. 在 `test_func3` 函数中，用到的栈空间为 200 字节，预留 200 字节的空间，刚刚好，和上面的两个函数不一样，这是为什么呢？