为什么我可以移动到eax，移动到ax，但是不能移动到al，甚至可能移动到ah？

Question

If I mov, eax 12345 and later mov var, eax (assuming var is a 32bit int etc..etc..) and output var later it'll output correctly.

Same with ax . mov ax, 65535 -> mov var16, ax will output correctly.

But if I mov into al or even ah it won't output anything.

Perhaps moving into ah not working makes sense. But why is that so for both?

typedef int int32;
typedef unsigned char int8;
typedef short int16;

int _tmain(int argc, _TCHAR* argv[])
{
    int8  c1 = 0;
    int8  c2 = 0;
    __asm
    {
        mov al, 255
        mov c1, al
    }

    cout << c1 << endl; //outputs nothing, same if I used ah

    //whereas

    int32 n = 0;
    __asm 
    {
        mov eax, 10000
        mov n, eax
    }

    cout << n << endl; // works fine, well, you get the picture

    return 0;
}

Answer 1

The problem is not (only) with your assembly code, it's the way you print your value:

cout << c1 << endl;

Even though c1 is unsigned char (aka uint8), C++ iostreams treat it the same way they treat an ordinary char. In other words, that code is roughly equivalent to:

printf("%c\n", c1);

Just cast it to unsigned int and it should work fine.

BTW, on some platforms (eg Linux PowerPC) the ordinary char is actually unsigned.

Answer 2

256 equals to 2^8

when you use int8 , it uses twos complement approach , so the number is between [-128 , 127]

I think if you change its type to unsigned integer , it must work