为什么编译器在堆栈中分配的比需要的多? [英] Why does the compiler allocate more than needed in the stack?

问题描述

我有一个简单的 C 程序.比方说，例如，我有一个 int 和一个长度为 20 的 char 数组.我总共需要 24 个字节.

I have a simple C program. Let's say, for example, I have an int and a char array of length 20. I need 24 bytes in total.

int main()
{
   char buffer[20];
   int x = 0;
   buffer[0] = 'a';
   buffer[19] = 'a';
}

堆栈需要与 16 字节边界对齐，因此我假设编译器将保留 32 字节.但是当我用 gcc x86-64 编译这样的程序并读取输出程序集时，编译器保留了 64 个字节.

The stack needs to be aligned to a 16 bytes boundary, so I presume a compiler will reserve 32 bytes. But when I compile such a program with gcc x86-64 and read the output assembly, the compiler reserves 64 bytes.

..gcc -S -o main.s main.c

给我:

    .file   "main.c"
    .def    __main; .scl    2;  .type   32; .endef
    .text
    .globl  main
    .def    main;   .scl    2;  .type   32; .endef
    .seh_proc   main
main:
    pushq   %rbp                        # RBP is pushed, so no need to reserve more for it
    .seh_pushreg    %rbp
    movq    %rsp, %rbp
    .seh_setframe   %rbp, 0
    subq    $64, %rsp                   # Reserving the 64 bytes
    .seh_stackalloc 64
    .seh_endprologue
    call    __main
    movl    $0, -4(%rbp)                # Using the first 4 bytes to store the int
    movb    $97, -32(%rbp)              # Using from RBP-32 
    movb    $97, -13(%rbp)              # to RBP-13 to store the char array
    movl    $0, %eax
    addq    $64, %rsp                   # Restoring the stack with the last 32 bytes unused
    popq    %rbp
    ret
    .seh_endproc
    .ident  "GCC: (x86_64-posix-seh-rev0, Built by MinGW-W64 project) 5.2.0"

这是为什么?当我编程汇编时，我总是只保留我需要的最小内存，没有任何问题.这是编译器的限制，无法评估所需的内存还是有原因?

Why is that? When I program assembly, I always reserve only the minimum memory I need without any problem. Is that a limitation of the compiler which has trouble evaluating the needed memory or is there a reason for that?

这里是 gcc -v

Using built-in specs.
COLLECT_GCC=gcc
COLLECT_LTO_WRAPPER=D:/Mingw64/bin/../libexec/gcc/x86_64-w64-mingw32/5.2.0/lto-wrapper.exe
Target: x86_64-w64-mingw32
Configured with: ../../../src/gcc-5.2.0/configure --host=x86_64-w64-mingw32 --build=x86_64-w64-mingw32 --target=x86_64-w64-mingw32 --prefix=/mingw64 --with-sysroot=/c/mingw520/x86_64-520-posix-seh-rt_v4-rev0/mingw64 --with-gxx-include-dir=/mingw64/x86_64-w64-mingw32/include/c++ --enable-shared --enable-static --disable-multilib --enable-languages=c,c++,fortran,objc,obj-c++,lto --enable-libstdcxx-time=yes --enable-threads=posix --enable-libgomp --enable-libatomic --enable-lto --enable-graphite --enable-checking=release --enable-fully-dynamic-string --enable-version-specific-runtime-libs --disable-isl-version-check --disable-libstdcxx-pch --disable-libstdcxx-debug --enable-bootstrap --disable-rpath --disable-win32-registry --disable-nls --disable-werror --disable-symvers --with-gnu-as --with-gnu-ld --with-arch=nocona --with-tune=core2 --with-libiconv --with-system-zlib --with-gmp=/c/mingw520/prerequisites/x86_64-w64-mingw32-static --with-mpfr=/c/mingw520/prerequisites/x86_64-w64-mingw32-static --with-mpc=/c/mingw520/prerequisites/x86_64-w64-mingw32-static --with-isl=/c/mingw520/prerequisites/x86_64-w64-mingw32-static --with-pkgversion='x86_64-posix-seh-rev0, Built by MinGW-W64 project' --with-bugurl=http://sourceforge.net/projects/mingw-w64 CFLAGS='-O2 -pipe -I/c/mingw520/x86_64-520-posix-seh-rt_v4-rev0/mingw64/opt/include -I/c/mingw520/prerequisites/x86_64-zlib-static/include -I/c/mingw520/prerequisites/x86_64-w64-mingw32-static/include' CXXFLAGS='-O2 -pipe -I/c/mingw520/x86_64-520-posix-seh-rt_v4-rev0/mingw64/opt/include -I/c/mingw520/prerequisites/x86_64-zlib-static/include -I/c/mingw520/prerequisites/x86_64-w64-mingw32-static/include' CPPFLAGS= LDFLAGS='-pipe -L/c/mingw520/x86_64-520-posix-seh-rt_v4-rev0/mingw64/opt/lib -L/c/mingw520/prerequisites/x86_64-zlib-static/lib -L/c/mingw520/prerequisites/x86_64-w64-mingw32-static/lib '
Thread model: posix
gcc version 5.2.0 (x86_64-posix-seh-rev0, Built by MinGW-W64 project) 

推荐答案

编译器可能确实为自己保留了额外的内存.

Compilers may indeed reserve additional memory for themselves.

Gcc 有一个标志，-mpreferred-stack-boundary，用于设置它将保持的对齐方式.根据文档，默认为 4，应该会产生 16 字节对齐，这是 SSE 指令所需要的.

Gcc has a flag, -mpreferred-stack-boundary, to set the alignment it will maintain. According to the documentation, the default is 4, which should produce 16-byte alignment, which needed for SSE instructions.

作为 VermillionAzure 在评论，你应该提供你的 gcc 版本和编译时选项(使用 gcc -v 来显示这些).

As VermillionAzure noted in a comment, you should provide your gcc version and compile-time options (use gcc -v to show these).

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