pre-increment and post-increment in printf

问题

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

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

And why is answer 4321 4321 ?

回答1:

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.

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);

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

回答2:

The code above is just broken. It is unspecified how it will work or what the answer will be. You need a sequence point between modifications or modifications and accesses.

来源：https://stackoverflow.com/questions/17030719/pre-increment-and-post-increment-in-printf