C ++编译器如何编译变量名？ [英] How does a C++ compiler compile variable names?

问题描述

我明白我没有说清楚。我认为，我的怀疑，可以总结在这：

在可执行文件（机器代码）如何表示变量？它们是静态内存地址吗？

用代码表示：

  int x = 5; 
 //一堆代码
 cin>> y; 
 cout<< x + 1; 
  

每个机器上的程序如何知道哪个地址将保存值5，保持输入的值，向现在保存的值添加1，最后打印相同的值。

- João

解决方案

这里是一个简单的程序在C：

  int main {
 int a = 5; 
 int b = 7; 
 
 int c = a + b; 
 
 return 0; 
} 
  

如果使用 gcc -m32 -S -O0 -o main.s main.c 在Linux下，你会得到类似这样的东西

  .filemain.c
 .text 
 .globl main 
 .type main，@function 
 main：
 .LFB0：
 / * ％ebp是基本指针寄存器* / 
 pushl％ebp 
 movl％esp，％ebp 
 
 / *这里我们为变量保留空间* / 
 subl $ 16，％esp 
 
 / * a的地址是％ebp  -  4 * / 
 movl $ 5，-4（％ebp）
 
 / * b的地址是％ ebp  -  8 * / 
 movl $ 7，-8（％ebp）
 
 / * a + b * / 
 movl -8（％ebp），％eax $ b b movl -4（％ebp），％edx 
 addl％edx，％eax 
 
 / * c的地址是％ebp  -  12 * / 
 movl％eax，-12 （％ebp）
 
 / * return 0 * / 
 movl $ 0，％eax 
 leave 
 ret 
  pre> 
 
 
可以看到，在这种情况下，变量的地址被计算为函数的基指针的偏移量。如果启用优化，变量的值可以存储在寄存器中。
 
I understand I did not make myself clear. My doubt, I think, could be summed up in this:

In an executable file(machine code) how are "variables" represented? Are they static memory addresses? Does the compiler gives each one a specific "name" (or just keeps the one you gave them)?

Expressed in code:
 int x=5;
 //Bunch of code
 cin>>y;
 cout<<x+1;
How does the program in each and every machine knows which address is going to hold the value 5, to hold the inputed value, to add 1 to the value it now holds and finally print that same value.

--João
解决方案
Here is a simple program in C:
int main() {
    int a = 5;
    int b = 7;

    int c = a + b;

    return 0;
}
If you compile it with gcc -m32 -S -O0 -o main.s main.c under Linux, you'll get something like this
    .file   "main.c"
    .text
    .globl  main
    .type   main, @function
main:
.LFB0:
    /* %ebp is a Base Pointer Register */
    pushl   %ebp
    movl    %esp, %ebp

    /* Here we reserve space for our variables */
    subl    $16, %esp

    /* a's address is %ebp - 4 */
    movl    $5, -4(%ebp)

    /* b's address is %ebp - 8 */
    movl    $7, -8(%ebp)

    /* a + b */
    movl    -8(%ebp), %eax
    movl    -4(%ebp), %edx
    addl    %edx, %eax

    /* c's address is %ebp - 12 */
    movl    %eax, -12(%ebp)

    /* return 0 */
    movl    $0, %eax
    leave
    ret
As you can see, in this case, variables' addresses are calculated as offsets of a base pointer of a function. If you enable optimisations, variables' values may be stored in registers.

                        
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