PowerPC内联汇编:将C值加载到寄存器中 [英] PowerPC Inline Assembly: Load C Value into Register

问题描述

使用GCC和inline assembly，我想将立即数加载到特定的寄存器r0中.但是，我没有得到正确的结果.

Using GCC and inline assembly, I want to load an immediate into a specific register r0. However, I'm not getting the right results.

unsigned short value = 0x1337;

asm volatile
(
"li 0, %0\n\t"
        "sc\n\t"
        "blr"
: /* Output registers */
:"r"(value) /* Input registers */
: /* No clobbered registers */
);

编译时会给出

li        r0, 9
sc
blr

9来自哪里?我想要指定的值0x1337代替. 这里是我看过的教程. >

Where does the 9 come from? I wanted the specified value 0x1337 instead. Here is a tutorial I looked at.

推荐答案

9是包含0x1337的寄存器，这正是您要的内容.注意value是输入寄存器吗? 9，又名r9，是一个完全有效的输入寄存器.这是我得到的程序集输出.

9 is the register containing 0x1337, which is exactly what you asked for. Notice how value is an input register? 9, a.k.a. r9, is a perfectly valid input register. This is the assembly output I get.

    li 9,4919
    li 0, 9
    sc
    blr

如果您想立即加载0x1337，只需使用它即可.

If you want to load 0x1337 as an immediate, just use that instead.

asm volatile (
    "li 0, 0x1337\n\t"
    "sc\n\t"
    "blr"
);

或者，只需使用"i"约束而不是"r"约束.

Or, just use the "i" constraint instead of the "r" constraint.

asm volatile (
    "li 0, %0\n\t"
    "sc\n\t"
    "blr"
    :
    : "i"(0x1337)
);

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