Question

我在这里做错了什么？

int val = 15;
asm ("movl %1, %%eax"::"r"(val):"%eax" );
__asm__ volatile ("int $0x80");

我想在eax中移动15然后调用中断

“错误”：'asm'：操作数超出范围

Answer 1

每个asm构造都是独立的，并且设置在一个中的值/寄存器与另一个没有连接。为了完成这项工作，您需要一个asm。此外，没有必要将值实际移动到eax中 - 这就是“a”输入约束的作用。所以你想要：

int val=15
asm volatile("int $0x80" : : "a"(val));

或只是

asm volatile("int $0x80"::"a"(15));

修改

各种约束字母的含义在the gcc documentation，但基本上，对于x86，它们是：

'r' -- any general register 'm' -- in memory addressable by an EA operand (base reg + index * scale + displacement) 'a' -- al/ax/eax/rax register (depending on the size of the operand) 'b' -- bl/bx/ebx/rbx register 'c' -- cl/cx/ecx/rcx register 'd' -- dl/dx/edx/rdx register 'A' -- edx:eax register pair (holding a 64-bit value) 'D' -- di/edi/rdi 'S' -- si/esi/rdi 'f' -- any 8087 fp register 't' -- ST(0) -- top of 8087 stack 'u' -- ST(1) -- second on 8087 stack 'y' -- any MMX register 'x' -- any XMM register

如果要在特定寄存器中放置多个内容，则需要多个输入，每个输入都有适当的约束。例如：

int read(int fd, void *buf, int size) { int rv; asm ("int $0x80" : "=a"(rv) : "a"(3), "b"(fd), "c"(buf), "d"(size) : "memory"); return rv; }

直接进行'读'系统调用。输入约束将各种参数放在eax / ebx / ecx / edx寄存器中，返回值最终在eax寄存器中。

对于与特定寄存器不对应的约束，您将在asm字符串中使用%n，它将被编译器选择的寄存器替换，但对于与特定寄存器对应的约束，没有必要直接提及它。

Answer 2

看起来他在以下代码示例中明确地将x放入EAX：

What is r() and double percent %% in GCC inline assembly language?

我使用g ++编译了该代码并通过gdb运行它在gdb中， info register eax 显示正确的数字

您使用的编译器和操作系统和架构是什么？如果您使用的是64位MSVS，则无法进行内联汇编。

#include<iostream>
using namespace std;

int main(int argc, char* argv[])
{

cout << "Compile via: g++ -O0 -ggdb -o asm.out asm.cpp" << endl;
cout << "Want to put 15 into register EAX and then call an interrupt." << endl;
cout << "gdb ./asm.out  --> stepi --> info register eaain." << endl;

int eaxin = 15, eaxout=0, ebxin=33, ebxout=0;
cout << "Before calling assembly routines:" << endl;
cout << "eaxin=" << eaxin << "\t" << "eaxout=" << eaxout << "\t" ;
cout << "ebxin=" << ebxin << "\t" << "ebxout=" << ebxout << endl;

asm ("movl %1, %%eax;"
     "movl %%eax, %0;"
    :"=r"(eaxout)   /* 1  eaxout is output operand 1 */
    :"r"(eaxin)     /* 0  eaxin is input operand 0*/
    :"%eax"
    );   /* %eax is clobbered register.  Why not two percent signs here? */

//Lets play in the ebx register.
asm ("movl %1, %%ebx;"
     "inc %%ebx;"
     "inc %%ebx;"
     "movl %%ebx, %0;"
    :"=r"(ebxout) 
    :"r"(ebxin) /* "+" is supposed to allow input and output but did not work */
    :"%ebx"
    ); 


//Call the system interrupt 128.
asm volatile ("int $0x80");

cout << "After calling assembly routines. Value of eaxout should have changed." << endl;
cout << "eaxin=" << eaxin << "  " << "eaxout=" << eaxout << "\t" ;
cout << "ebxin=" << ebxin << "\t" << "ebxout=" << ebxout << endl;
cout << endl;
cout << "https://stackoverflow.com/questions/3589157/need-some-help-understanding-gcc-inline-assembly-language?rq=1" << endl;
cout << "Note: inline assembly CANNOT be done with MS VisualStudio 64bit." << endl;
cout << "   option1:   Use a separate file for the assembler code." << endl;
cout << "   option2:   Use instrinsics." << endl;
cout << " Look for intrn.h at http://software.intel.com/en-us    /node/181178?wapkw=ain86_64+assembly "  << endl;
return 0;
}

（gdb）信息注册表eax ebx
     eax 0xf 15
     ebx 0x0 0
   （gdb）步骤
     35 asm volatile（“int $ 0x80”）;
   （gdb）信息注册表eax ebx
     eax 0x23 35
     ebx 0x23 35

c变量存储在eax寄存器中

2 个答案: