C90 printf with \n or \r\n not working in cygwin; but fflush(stdout); works fine. why?

Question

Cygwin64 bit

Command to compile:

gcc hello.c -o hello -ansi -pedantic-errors

Command to run

./hello

hello.c

#include<stdio.h>
int main() {
    /*setbuf(stdout, 0);        I KNOW THIS WILL WORK IF ADDED, it is a solution, but I want to know why line break itself is not working*/
    printf("hello world!\n");
    printf("hello world again!\r\n");
    /*fflush(stdout);           without fflush, the above strings are not showing*/

    while(1)
    {
    }
}

Questions:

1. I don't want a fflush after every printf to let the terminal show the string in time, then how?

   ANSWER: setbuf(stdout, 0);

2. Why is "\n" or "\r\n" not working in my case considering lots of posts pointed out a line break will fix the problem?

3. Is it true that cygwin's terminal behaves differently than normal Linux's terminal? Since I don't have linux installed, anyone give a test for me?

4. Or let me ask a more general question: On which kinds of terminals, the sentence "a new line will force flush" is true?

Thanks

Answer 1

It seems that in Cygwin, stdout isn't identified as a terminal (but as a pipe), so it isn't line-buffered by default.

Based on this answer, perhaps linking with the Cygwin DLL would help.

Answer 2

I know nothing about cgywin. but here I do a test in Linux.

I try the code below, and compile by : gcc -std=c90 filename.c

Without fflush It print all words before loop,so I think the newline flush the buffer! It just work!

I use SUSE gcc .

#include<stdio.h>
int main() {
    /*setbuf(stdout, 0);        I KNOW THIS WILL WORK IF ADDED, it is a solution, but I want to know why line break itself is not working*/
    printf("hello world!\n"); /*without fflush, not shown*/
    printf("hello world again!\r\n"); /*without fflush, not shown*/
  /* fflush(stdout);*/

    while(1)
    {
    }
}

Answer 3

That program works for me under 32-bit Cygwin. Specifically, when I compile and execute this program:

#include<stdio.h>
int main() {
    /*setbuf(stdout, 0);        I KNOW THIS WILL WORK IF ADDED, it is a solution, but I want to know why line break itself is not working*/
    printf("hello world!\n");
    printf("hello world again!\r\n");
    /*fflush(stdout);           without fflush, the above strings are not showing*/

    while(1)
    {
    }
}

it produces this output:

hello world!
hello world again!

and then hangs until I kill it with Ctrl-C.

I get the same behavior invoking the program from bash under the Windows console, mintty, and xterm (I doubt that the terminal would make any difference).

I'm using 32-bit Cygwin under 64-bit Windows 7. You say you're using 64-bit Cygwin, which was just announced a few days ago; I haven't tried it yet.

I suspect an issue with 64-bit vs. 32-bit Cygwin. I suggest you post to the Cygwin mailing list.

Here's a cleaned-up version of your program that should exhibit the same issue (please verify that the comments are correct):

#include <stdio.h>
int main(void) {
    /* Adding setbuf(stdout, 0) here makes the output appear */
    printf("hello world!\n");
    /* Adding fflush(stdout) here makes the output appear */
    while(1) {
        /* nothing */
    }
}

Answer 4

I've just had this problem. I've got some code I've been working on a while and had to reinstall Cygwin on my PC. I installed the 64-bit version this time, it was x86 previously.

Built my code, ran it in the bash shell and no output.

What I didn't realize is that I have a Cygwin1.dll from the previous 32-bit version of cygwin still in my bin folder. I renamed this and ran my program and the output worked.

You may have a different issue, but if you've got different dll versions in use your bin other folders, it may cause this problem.