2.4 OpenEuler中C语言中的函数调用测试

任务详情

1. 在X86_64架构下实践2.4中的内容
2. 通过GDB查看寄存器的内容，把教材中的图填入具体的值
3. 把2.4的C代码在OpenEuler中重新实践一遍，绘制出ARM64的逻辑框图

任务一X86_64架构下实践2.4

操作环境: Ubuntu Kylin

2.4.1

code

#include <stdio.h>

int sub(int x,int y)
{
    int u,v;
    u = 4;v = 5;
    return x+y+u+v;
}

int main()
{
    int a,b,c;
    a = 1;b = 2;c = 3;
    c = sub(a,b);
    printf("c = %d\n",c);
}

code.s

sub:
.LFB0:
        .cfi_startproc
        endbr64
        pushq   %rbp
        .cfi_def_cfa_offset 16
        .cfi_offset 6, -16
        movq    %rsp, %rbp
        .cfi_def_cfa_register 6
        movl    %edi, -20(%rbp)
        movl    %esi, -24(%rbp)
        movl    $4, -8(%rbp)
        movl    $5, -4(%rbp)
        movl    -20(%rbp), %edx
        movl    -24(%rbp), %eax
        addl    %eax, %edx
        movl    -8(%rbp), %eax
        addl    %eax, %edx
        movl    -4(%rbp), %eax
        addl    %edx, %eax
        popq    %rbp
        .cfi_def_cfa 7, 8
        ret
        .cfi_endproc
.LFE0:
        .size   sub, .-sub
        .section        .rodata
.LC0:
        .string "c = %d\n"
        .text
        .globl  main
        .type   main, @function
main:
.LFB1:
        .cfi_startproc
        endbr64
        pushq   %rbp
        .cfi_def_cfa_offset 16
        .cfi_offset 6, -16
        movq    %rsp, %rbp
        .cfi_def_cfa_register 6
        subq    $16, %rsp
        movl    $1, -12(%rbp)
        movl    $2, -8(%rbp)
        movl    $3, -4(%rbp)
        movl    -8(%rbp), %edx
        movl    -12(%rbp), %eax
        movl    %edx, %esi
        movl    %eax, %edi
        call    sub
        movl    %eax, -4(%rbp)
        movl    -4(%rbp), %eax
        movl    %eax, %esi
        movl    $.LC0, %edi
        movl    $0, %eax
        call    printf
        movl    $0, %eax
        leave
        .cfi_def_cfa 7, 8
        ret
        .cfi_endproc

GDB查看寄存器内容

填写教材中内容

HIGH	PC	rbp	c	b	a	tmp	pc	rbp	v	u	tmp	x	y	LOW
....	unknown	0x0	0x3	0x2	0x1	4bytes	0x401191	0x7fffffffe240	0x5	0x4		0x1	0x2	....

任务二arm64下实践2.4

操作环境: openeuler with arm