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
997 views
in Technique[技术] by (71.8m points)

c - Why is the line following printf(), a call to sleep(), executed before anything is printed?

I thought I was doing something simple here, but C decided to go asynchronous on me. I'm not sure what's going on. Here's my code:

#include <stdio.h>
int main() {
    printf("start");
    sleep(5);
    printf("stop");
}

When I compile and run, I notice that sleep(5) works like a charm. But the compiler decided it was a good idea to skip the first printf() and go out of order, so when running, the program waits for 5 seconds and then prints startstop.

What's the deal? My theory is that the program initiates the print operation with the shell, then continues with the program, leaving Bash to wait until the program is no longer busy to actually render the strings. But I really don't know.

Thanks

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)

printf uses buffered output. This means that data first accumulates in a memory buffer before it is flushed to the output source, which in this case is stdout (which generally defaults to console output). Use fflush after your first printf statement to force it to flush the buffered data to the output source.

#include <stdio.h>
int main() {
    printf("start");
    fflush(stdout);
    sleep(5);
    printf("stop");
}


Also see Why does printf not flush after the call unless a newline is in the format string?


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

...