C变量的内容复制到寄存器（GCC） [英] Copy content of C variable into a register (GCC)

问题描述

由于我很新的GCC，我面临的内联汇编code的一个问题。问题是，我无法弄清楚如何复制C变量（其类型为 UINT32 ）的进入寄存器的内容 EAX 。我曾尝试低于code：

Since I'm very new to GCC, I'm facing a problem in inline assembly code. The problem is that I'm not able to figure out how to copy the contents of a C variable (which is of type UINT32) into the register eax. I have tried the below code:

__asm__
(
    // If the LSB of src is a 0, use ~src.  Otherwise, use src.
    "mov     $src1, %eax;"
    "and     $1,%eax;"
    "dec     %eax;"
    "xor     $src2,%eax;"

    // Find the number of zeros before the most significant one.
    "mov     $0x3F,%ecx;"
    "bsr     %eax, %eax;"
    "cmove   %ecx, %eax;"
    "xor     $0x1F,%eax;"
);

但 MOV $ SRC1，EAX％; 不起作用

有人能提出一个解决的办法？

Could someone suggest a solution to this?

推荐答案

我猜测你正在寻找的是的扩展组件例如：

I guess what you are looking for is extended assembly e.g.:

    int a=10, b;
    asm ("movl %1, %%eax;   /* eax = a */
          movl %%eax, %0;" /* b = eax */
         :"=r"(b)         /* output */
         :"r"(a)         /* input */
         :"%eax"        /* clobbered register */
         );        

在上面的例子中，我们做了的值b 等于的 A 使用汇编指令和 EAX 注册：

In the example above, we made the value of b equal to that of a using assembly instructions and eax register:

int a = 10, b;
b = a;

请参阅内嵌批注。

注意：

mov $4, %eax          // AT&T notation

mov eax, 4            // Intel notation

有关GCC 内联汇编一个良好的阅读环境。

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