GCC 内存对齐

在学习C 汇编中发现了一个奇怪的现象,如下

    int foo1(int m,int n)    {        int p=m*n;        return p;    }    int foo(int a,int b)    {        int c=a+1;        int d=b+2;        int e=foo1(c,d);        return e;    }    int main()    {        int result=foo(3,4);        return 0;    }

生成的汇编代码如下

    .file   "foo.c"    .text    .globl  foo1    .type   foo1, @functionfoo1:.LFB0:    .cfi_startproc    pushl   %ebp    .cfi_def_cfa_offset 8    .cfi_offset 5, -8    movl    %esp, %ebp    .cfi_def_cfa_register 5    subl    $16, %esp    movl    8(%ebp), %eax    imull   12(%ebp), %eax    movl    %eax, -4(%ebp)    movl    -4(%ebp), %eax    leave    .cfi_restore 5    .cfi_def_cfa 4, 4    ret    .cfi_endproc.LFE0:    .size   foo1, .-foo1    .globl  foo    .type   foo, @functionfoo:.LFB1:    .cfi_startproc    pushl   %ebp    .cfi_def_cfa_offset 8    .cfi_offset 5, -8    movl    %esp, %ebp    .cfi_def_cfa_register 5    subl    $16, %esp    movl    8(%ebp), %eax    addl    $1, %eax    movl    %eax, -12(%ebp)    movl    12(%ebp), %eax    addl    $2, %eax    movl    %eax, -8(%ebp)    pushl   -8(%ebp)    pushl   -12(%ebp)    call    foo1    addl    $8, %esp    movl    %eax, -4(%ebp)    movl    -4(%ebp), %eax    leave    .cfi_restore 5    .cfi_def_cfa 4, 4    ret    .cfi_endproc.LFE1:    .size   foo, .-foo    .globl  main    .type   main, @functionmain:.LFB2:    .cfi_startproc    pushl   %ebp    .cfi_def_cfa_offset 8    .cfi_offset 5, -8    movl    %esp, %ebp    .cfi_def_cfa_register 5    subl    $16, %esp    pushl   $4    pushl   $3    call    foo    addl    $8, %esp    movl    %eax, -4(%ebp)    movl    $0, %eax    leave    .cfi_restore 5    .cfi_def_cfa 4, 4    ret    .cfi_endproc.LFE2:    .size   main, .-main    .ident  "GCC: (Ubuntu 5.4.0-6ubuntu1~16.04.5) 5.4.0 20160609"    .section    .note.GNU-stack,"",@progbits

代码使用32位编译的，一个int的字节是4
把上面的帧栈画出来

帧栈 ebp(main) 4 (para b in foo) 3 (para a in foo) ret(return of foo) ebp(foo) s[fool return value](local var in foo) d(6)[b+2](local var in foo) c(4)[a+1](local var in foo) d(para var in foo1) c(para var in foo1) ret(return of foo1) ebp(foo1就是这里) p(laocao var in foo1) null null null

汇编中有一行

subl $16 %esp

也就是开辟了 四”块”栈区，但实际上，foo1函数中只有一个local variable —– p

后询问大佬得知了如标题，同时又认识了一本book：）

图片出自《深入理解计算机系统》