Clang 填充数组

发布于 2025-01-16 21:14:33 字数 2195 浏览 2 评论 0原文

我试图了解函数调用在机器级别是如何工作的。为此，我创建了一个像这样的虚拟 C++ 函数：

int func(int a[], int n) {
    char arr[n];
    arr[n - 99] = 100; //to prevent compiler from optimizing my function to noop
    return arr[12]; //to prevent compiler from optimizing my function to noop
}

并且，在使用 x86-64 clang 13.0.1 编译时，我得到了以下程序集

push    rbp
mov     rbp, rsp
mov     eax, esi
mov     rcx, rsp
add     rax, 15 // Instruction 1
and     rax, -16 // Instruction. 2
mov     rdx, rcx
sub     rdx, rax
mov     rsp, rdx
movsxd  rsi, esi
mov     byte ptr [rsi + rdx - 99], 100
neg     rax
movsx   eax, byte ptr [rcx + rax + 12]
mov     rsp, rbp
pop     rbp
ret

指令 1 和 2 正在从变量 n 计算数组 arr 的大小。计算出的大小是 16 的下一个最大倍数 n。例如，即使n = 1，clang 也会分配16 字节的内存。我很困惑为什么 clang 会尝试将数组大小设置为 16 字节的倍数？

我检查了 arr 的不同数据类型，但它们都有相同的行为。因此，我认为这不是出于对齐目的，因为不同的数据类型将具有不同的对齐大小。

我也检查了不同的编译器。所有分配的数组的大小都是 16 的倍数。我缺少一些基本的东西。

<一href="https://godbolt.org/#g:!((g:!((g:!((h:codeEditor,i:(文件名:%271%27,fontScale:14,fontUsePx:%270% 27,j:1,lang:c%2B%2B,选择:(endColumn:2,endL ineNumber:5,positionColumn:2,positionLineNumber:5,selectionStartColumn:1,selectionStartLineNumber:1,startColumn:1,startLineNumber:1),来源:%27int+fun c(int+a%5B%5D,+int+n)+%7B%0A%09char+arr%5Bn%5D%3B%0A++++arr%5Bn+-+99%5D+%3D +100%3B%0A++++返回+arr%5B12%5D%3B%0A%7D%27),l:%275%27,n:%270%27,o:%27C%2B %2B+来源+%231%27,t:%270%27)),k:50,l:%274%27,n:%270%27,o:%27%27,s:0,t:% 270 %27),(g:!((h:编译器,i:(编译器:clang1301,过滤器:(b:%270%27,二进制:%271%2 7、仅注释:%270%27、分解:%270%27、指令:%270%27、执行:%271%27、英特尔:%270%27、库代码:%270%27、修剪:%271%27 ),flagsViewOpen:%271%27,fontSc ale:14,fontUsePx:%270%27,j:1,lang:c%2B%2B,库:!(),选项:%27-O3%27,选择:(endColumn:1,endLineNumber:1,positionColumn :1,位置行号:1,选择ctionStartColumn:1,selectionStartLineNumber:1,startColumn:1,startLineNumber:1),源:1,树:%271%27),l:%275%27,n:%270%27,o:%27x86-64+铿锵+13.0.1+(C% 2B%2B,+编辑器+%231,+编译器+%231)%27,t:%270%27)),k:50,l:%274%27,n:%270%27, o:%27%27,s:0,t:%270%27)),l:%272%27,n:%270%27,o:%27%27,t:%270%27)),版本：4" rel="nofollow noreferrer">这里是作为参考的链接。

原文

I am trying to understand how function calling works at machine level. For that I created a dummy C++ function like this :

int func(int a[], int n) {
    char arr[n];
    arr[n - 99] = 100; //to prevent compiler from optimizing my function to noop
    return arr[12]; //to prevent compiler from optimizing my function to noop
}

And, I got following assembly when compiling with x86-64 clang 13.0.1

push    rbp
mov     rbp, rsp
mov     eax, esi
mov     rcx, rsp
add     rax, 15 // Instruction 1
and     rax, -16 // Instruction. 2
mov     rdx, rcx
sub     rdx, rax
mov     rsp, rdx
movsxd  rsi, esi
mov     byte ptr [rsi + rdx - 99], 100
neg     rax
movsx   eax, byte ptr [rcx + rax + 12]
mov     rsp, rbp
pop     rbp
ret

Instructions 1 and 2 are calculating size of array arr from variable n. The calculated size is next largest multiple of 16 than n. For example, even if n = 1, clang is allocating 16 bytes of memory. I am confused as to why would clang try to make array sizes as multiple of 16 bytes?

I checked with different data types for arr but all have same behavior. So, I don't think this is for alignment purposes as different data type would have different alignment size.

I checked with different compilers as well. All are allocating arrays of sizes multiple of 16. There is something fundamental that I am missing.

Here is the link as reference.

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