Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
183 views
in Technique[技术] by (71.8m points)

c - pre-increment and post-increment in printf

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 ?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

1.4m articles

1.4m replys

5 comments

56.9k users

...