linux - 将内存分配给递归函数

Question

我写了一个简单的程序，如下所示。

#include<stdio.h>
int foo(int i)
{
    int k=9;
    if(i==10)
            return 1;
    else
            foo(++i);
    open("1",1);
}
int main()
{
    foo(1);
}

我这样做的目的是检查如何为堆栈上的函数中的变量(在本例中为 int k)分配内存。我使用一个开放的系统调用作为标记。 strace 的输出如下:

execve("./a.out", ["./a.out"], [/* 25 vars */]) = 0
brk(0)                                  = 0x8653000
access("/etc/ld.so.nohwcap", F_OK)      = -1 ENOENT (No such file or            directory)
mmap2(NULL, 8192, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) =     0xb777e000
access("/etc/ld.so.preload", R_OK)      = -1 ENOENT (No such file or directory)
open("/etc/ld.so.cache", O_RDONLY|O_CLOEXEC) = 3
fstat64(3, {st_mode=S_IFREG|0644, st_size=95172, ...}) = 0
mmap2(NULL, 95172, PROT_READ, MAP_PRIVATE, 3, 0) = 0xb7766000
close(3)                                = 0
access("/etc/ld.so.nohwcap", F_OK)      = -1 ENOENT (No such file or     directory)
open("/lib/i386-linux-gnu/libc.so.6", O_RDONLY|O_CLOEXEC) = 3
read(3, "\177ELF\1\1\1\0\0\0\0\0\0\0\0\0\3\0\3\0\1\0\0\0000\226\1\0004\0\0\0"..., 512) = 512
fstat64(3, {st_mode=S_IFREG|0755, st_size=1734120, ...}) = 0
mmap2(NULL, 1743580, PROT_READ|PROT_EXEC, MAP_PRIVATE|MAP_DENYWRITE, 3, 0) =     0xb75bc000
mmap2(0xb7760000, 12288, PROT_READ|PROT_WRITE,     MAP_PRIVATE|MAP_FIXED|MAP_DENYWRITE, 3, 0x1a4) = 0xb7760000
mmap2(0xb7763000, 10972, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_FIXED|MAP_ANONYMOUS, -1, 0) = 0xb7763000
close(3)                                = 0
mmap2(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) =     0xb75bb000
set_thread_area({entry_number:-1 -> 6, base_addr:0xb75bb900, limit:1048575, seg_32bit:1, contents:0, read_exec_only:0, limit_in_pages:1, seg_not_present:0, useable:1}) = 0
mprotect(0xb7760000, 8192, PROT_READ)   = 0
mprotect(0x8049000, 4096, PROT_READ)    = 0
mprotect(0xb77a1000, 4096, PROT_READ)   = 0
munmap(0xb7766000, 95172)               = 0
open("1", O_WRONLY)                     = -1 ENOENT (No such file or     directory)
open("1", O_WRONLY)                     = -1 ENOENT (No such file or     directory)
open("1", O_WRONLY)                     = -1 ENOENT (No such file or directory)
open("1", O_WRONLY)                     = -1 ENOENT (No such file or     directory)
open("1", O_WRONLY)                     = -1 ENOENT (No such file or     directory)
open("1", O_WRONLY)                     = -1 ENOENT (No such file or     directory)
open("1", O_WRONLY)                     = -1 ENOENT (No such file or     directory)
open("1", O_WRONLY)                     = -1 ENOENT (No such file or directory)
open("1", O_WRONLY)                     = -1 ENOENT (No such file or directory)
exit_group(-1)                          = ?

在 strace 输出的末尾，您可以看到在打开的系统调用之间没有调用任何系统调用。那么，在没有系统调用的情况下，如何将内存分配到堆栈上以供调用的函数使用？

Answer 1

主线程的堆栈内存由内核在 execve() 期间分配。系统调用。在此调用期间，还设置了可执行文件中定义的其他映射(并且可能还用于可执行文件中指定的动态链接器)。对于 ELF 文件，这是在 fs/binfmt_elf.c 中完成的.

其他线程的栈内存是mmap()由线程支持库编辑，它通常是 C 运行时库的一部分。

您还应该注意，在虚拟内存系统上，主线程堆栈由内核增长以响应页面错误，直到达到可配置的限制(由 ulimit -s 显示)。