在SPARC装配参数的printf格式说明？ [英] printf format specifier in SPARC assembly argument?

问题描述

如何让这款C与SPARC汇编等价的：

How do you get the equivalent of this C with SPARC assembly:

printf( "Hello, my name is %s.\n", name );

使用函数原型：

void printName( const char* msg, const char* name )

，其中味精是你好，我的名字为％s \\ N。

where msg is "Hello, my name is %s.\n".

我知道我可以定义你好，我的名字为％s \\ N与.asciz数据段，而不必第一个参数味精，但有没有办法将字符串传递到汇编函数那会在它％的标识符？可一个char *甚至采取在格式标识符？我试过以下，但我得到一个核心转储。

I know that I can define "Hello, my name is %s.\n" in the data segment with .asciz without having the first argument msg, but is there a way to pass a string into an assembly function that would have a %s identifier in it? Can a char* even take in a format identifier? I've tried the following but I get a core dump.

用C 函数调用：

char * msg = "Hello, my name is %s.\n";
char * name = "Foo";

printName( msg, name );

大会：

mov %i0, %o0
mov %i1, %o1
call printf, 2
nop

也许我不能正确地接近函数原型？

Maybe I'm not approaching the function prototype correctly?

推荐答案

我其实不安静知道你做错了，但下面的程序工作，因为它应该：

I am actually not quiet sure what you are doing wrong, but the following program works as it should:

        .data
s0:     .asciz  "foo %s\n"
s1:     .asciz  "bar"
        .text
        .global main
main:
        save    %sp, -96, %sp
        set     s0, %o0
        set     s1, %o1
        call    prtnam
        nop
        ret
        restore
prtnam:
        save    %sp, -96, %sp
        mov     %i0, %o0
        call    printf
        mov     %i1, %o1
        ret
        restore

关于参数的传递，少数像这样在寄存器中传递的参数。

Regarding the passing of arguments, a small number of arguments like this are passed in registers.

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