如何在不知道的情况下更改变量的值？

发布于 2025-01-18 19:02:16 字数 901 浏览 4 评论 0原文

我想通过这个方法来验证一下 volatile 的作用。但我的内联汇编代码似乎无法在编译器不知道的情况下修改 i 的值。根据我看的文章，我只需要写像 __asm { mov dword ptr [ebp-4], 20h } 这样的汇编代码，我想我写的和他做的一样。

实际输出：

before = 10
after = 123

预期输出：

before = 10
after = 10

文章链接： https://www.runoob.com /w3cnote/c-volatile-keyword.html

#include <stdio.h>

int main() {
    int a, b;
    // volatile int i = 10;
    int i = 10;

    a = i;
    printf("before = %d\n", a);

    // Change the value of i in memory without letting the compiler know.
    // I can't run the following statement here, so I wrote one myself
    // mov dword ptr [ebp-4], 20h
    asm("movl $123, -12(%rbp)");

    b = i;
    printf("after = %d\n", b);
}

原文

I want to verify the role of volatile by this method. But my inline assembly code doesn't seem to be able to modify the value of i without the compiler knowing. According to the articles I read, I only need to write assembly code like __asm { mov dword ptr [ebp-4], 20h }, I think I write the same as what he did.

actual output:

before = 10
after = 123

expected output:

before = 10
after = 10

Article link: https://www.runoob.com/w3cnote/c-volatile-keyword.html

#include <stdio.h>

int main() {
    int a, b;
    // volatile int i = 10;
    int i = 10;

    a = i;
    printf("before = %d\n", a);

    // Change the value of i in memory without letting the compiler know.
    // I can't run the following statement here, so I wrote one myself
    // mov dword ptr [ebp-4], 20h
    asm("movl $123, -12(%rbp)");

    b = i;
    printf("after = %d\n", b);
}

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我喜欢麦丽素 2025-01-25 19:02:16

我想验证volatile ...
的作用

您不能。

如果变量不是挥发性，则编译器 May Optimize;它确实不需要来做到这一点。

编译器可以始终将任何变量视为volatile。

如何在不知道的情况下更改变量的值？

创建第二个线程写入变量。

示例

以下示例用于Linux（在Windows下，您需要一个与pthread_create（））不同的功能：

#include <stdio.h>
#include <pthread.h>

int testVar;
volatile int waitVar;

void * otherThread(void * dummy)
{
    while(waitVar != 2) { /* Wait */ }
    testVar = 123;
    waitVar = 3;
    return NULL;
}

int main()
{
    pthread_t pt;
    waitVar = 1;
    pthread_create(&pt, 0, otherThread, NULL);
    testVar = 10;
    waitVar = 2;
    while(waitVar != 3) { /* Wait */ }
    printf("%d\n", testVar - 10);
    return 0;
}

如果您使用gcc -o0 -ox XC-编译lpthread，编译器不会优化，并且像所有变量一样volatile。 printf（）打印113。

如果您使用-O3而不是-O0，printf（）打印0 。

/-O3）。

（使用GCC 9.4.0编译器进行测试。）

I want to verify the role of volatile ...

You can't.

If a variable is not volatile, the compiler may optimize; it does not need to do this.

A compiler may always treat any variable as volatile.

How to change the value of a variable without the compiler knowing?

Create a second thread writing to the variable.

Example

The following example is for Linux (under Windows, you need a different function than pthread_create()):

#include <stdio.h>
#include <pthread.h>

int testVar;
volatile int waitVar;

void * otherThread(void * dummy)
{
    while(waitVar != 2) { /* Wait */ }
    testVar = 123;
    waitVar = 3;
    return NULL;
}

int main()
{
    pthread_t pt;
    waitVar = 1;
    pthread_create(&pt, 0, otherThread, NULL);
    testVar = 10;
    waitVar = 2;
    while(waitVar != 3) { /* Wait */ }
    printf("%d\n", testVar - 10);
    return 0;
}

If you compile with gcc -O0 -o x x.c -lpthread, the compiler does not optimize and works like all variables are volatile. printf() prints 113.

If you compile with -O3 instead of -O0, printf() prints 0.

If you replace int testVar by volatile int testVar, printf() always prints 113 (independent of -O0/-O3).

(Tested with the GCC 9.4.0 compiler.)

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