printf中的前增量和后增量 [英] pre-increment and post-increment in printf

问题描述

int main()
{
    int value = 4321;
    int *ptrVal = &value;
    printf("%d %d",++value,(*(int*)ptrVal)--);
    return 0;
}

上述打印语句中的预增/后增如何工作?

How does pre-increment/post increment works in above print statement ?

为什么回答4321 4321?

And why is answer 4321 4321 ?

推荐答案

您要在两个序列点之间两次修改对象value:您正在调用未定义的行为.未定义的行为意味着您的程序可以打印4321 4321，打印42甚至崩溃.

You are modifying the object value twice between two sequence points: you are invoking undefined behavior. Undefined behavior means your program can print 4321 4321, print 42 or even just crash.

您的程序的正确版本为:

A correct version of your program would be:

int value = 4321;
int *ptrVal = &value;

++value;
(*ptrVal)--;  // no need to cast to int *

printf("%d %d", value, *ptrVal);  // same as printf("%d %d", value, value);

当然，您不需要任何临时指针即可实现这一目标.

Of course you don't need any temporary pointer to achieve this.

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