C中函数调用中后缀或前缀递增的未定义行为 [英] Undefined Behavior of Postfix or Prefix Increment in Function Calls in C

问题描述

我在本网站看到，函数调用中的前缀增量或后缀增量可能导致未定义的行为。我最近经历了其中一次。源代码如下所示：

I have seen in this site that prefix increment or postfix increment in a function call may cause undefined behavior. I have gone through one of those recently. The source code is something like this :

#include <stdio.h>

void call(int,int,int);
int main()
{
    int a=10;
    call(a,a++,++a);
    printf("****%d %d %d***_\n",a,a++,++a);
    return 0;
}

void call(int x,int y,int z)
{
    printf("%d %d %d",x,y,z);
}

输出结果为12 11 12 **** 14 13 14 * ** _。但是，当在功能中首先打印a时，它不应该是10吗？为什么它变成了12？另外，为什么++从12减少到11？有人可以请解释一下吗？谢谢。

The output comes out as 12 11 12****14 13 14***_. But, when a is printed first in the function, shouldn't it be 10? Why does it become 12? Also, why does a++ decrease from 12 to 11? Can someone please kindly explain? Thank you.

推荐答案

这是未定义的行为，因此完全取决于编译器的执行顺序操作完成：

That is undefined behaviour and as such it is entirely up to the implementation of the compiler in which order the following operations are done:

1. 提交参数 a


2. 提交参数 a ++


3. 提交参数 ++ a li>
   
4. 增量 a ++ a


5. 为 a ++
增加 a

1. submit argument a
2. submit argument a++
3. submit argument ++a
4. increment a for ++a
5. increment a for a++

编译器知道的唯一事情是：2.必须在5和4之前发生。

The only thing that the compiler knows is: 2. has to happen before 5. and 4. has to happen before 3.

您正在观察：

You are observing:

++a;
submit argument 2
a++;
submit the other arguments

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