Question

我有一个C代码，用于调用ARM Assembly中定义的函数。必须传递两个参数。

如果函数调用如下所示：

int main()
{
    try {
        std::ifstream text_file;
        ics::safe_open(text_file,"Enter file name to analyze","C:\\Users\\Xari\\Downloads\\program1\\graph1.txt");
        Graph graph = read_graph(text_file);

        print_graph(graph);

        while (true) {
            std::string input;
            std::cout << "Enter the name of a starting node (enter quit to quit) D: ";
            std::cin >> input;
            std::cout << "INVISIBLE DEBUGGER MESSAGE" << std::endl;
            if (input != "quit")
                std::cout << "Reachable from node name " << input << " = " << reachable(graph, input) << std::endl;
            else
                break;
        }

    } catch (ics::IcsError& e) {
        std::cout << e.what() << std::endl;
    }

    return 0;
}

寄存器functionName(a, b)和x0按哪个顺序保存这些值？它是x1持有x0而a持有x1还是相反？

Answer 1

提出这个问题比花时间更长。

extern void bar ( unsigned int, unsigned int );

void foo ( void )
{
    bar(5,7);
}

编译然后反汇编

传统手臂

00000000 <foo>:
   0:   e3a00005    mov r0, #5
   4:   e3a01007    mov r1, #7
   8:   eafffffe    b   0 <bar>

aarch64

0000000000000000 <foo>:
   0:   528000e1    mov w1, #0x7                    // #7
   4:   528000a0    mov w0, #0x5                    // #5
   8:   14000000    b   0 <bar>
   c:   d503201f    nop

MSP430

00000000 <foo>:
   0:   3e 40 07 00     mov #7, r14 ;#0x0007
   4:   3f 40 05 00     mov #5, r15 ;#0x0005
   8:   b0 12 00 00     call    #0x0000 
   c:   30 41           ret

PDP-11

00000000 <_foo>:
   0:   1166            mov r5, -(sp)
   2:   1185            mov sp, r5
   4:   15e6 0007       mov $7, -(sp)
   8:   15e6 0005       mov $5, -(sp)
   c:   09f7 fff0       jsr pc, 0 <_foo>
  10:   65c6 0004       add $4, sp
  14:   1585            mov (sp)+, r5
  16:   0087            rts pc

从ARM程序集到C的函数调用的参数传递约定

1 个答案: