I am using printf("%d", 15); and nothing prints on the console.
I tried calling setvbuf (stdout, NULL, _IONBF, 0); first, nothing changed.
Any ideas how to tackle this issue ?
printf buffers the output. It will not flush the buffer (i.e. actually write out the contents) until a newline is reached.
The best remedy is to use printf("%d\n", 15);. Alternatively you can flush the buffer using fflush(stdout);
You can suppress the buffering behaviour by writing setbuf(stdout, NULL); but I wouldn't recommend your interfering with the workings in that way.
Related
I want to turn off the buffering for the stdout for getting the exact result for the following code
while(1) {
printf(".");
sleep(1);
}
The code printf bunch of '.' only when buffer gets filled.
You can use the setvbuf function:
setvbuf(stdout, NULL, _IONBF, 0);
Here're some other links to the function.
POSIX
C/C++
You can also use setbuf
setbuf(stdout, NULL);
This will take care of everything
Use fflush(FILE *stream) with stdout as the parameter.
http://www.elook.org/programming/c/fflush.html
You can do this:
write(1, ".", 1);
instead of this:
printf(".");
Use fflush(stdout). You can use it after every printf call to force the buffer to flush.
When I run the code below, it runs the system command on the 4th last line before displaying the string 'proceeding' before it! I'm wondering why and how to fix it, any ideas?
if ((strlen(command)>0) && (command[strlen (command) - 1] == '\n'))
command[strlen (command) - 1] = '\0';
printf("proceeding"); // <-- the string
strcat(command,contents);
strcat(command,subject);
system(command); // <-- offending system command
sleep (1);
printf("\n ----- Search complete for: [%s]",command);
getchar();
There are of course variables such as 'command' and 'subject' which are manipulated and declared outside the code above, so If you need context than I will post the rest of the source code below.
Pull the chain and flush:
I.e.
After
printf("proceeding");
Put
fflush(stdout);
That will flush the stuff in the buffer (bowl!)
Before the system command is executed.
Try adding a '\n' to the printf.
It forces flushing the printf buffer. Otherwise, it is not necessary that printf immediately prints the passed params. You can google flushing the buffer latter
stdout is line buffered, so it will only display what's in the buffer after it reaches a newline
change printf("proceeding"); to printf("proceeding\n"); or flush stdout with fflush(stdout); will do the work.
The stdio stdout stream is line or fully buffered by default. To set it unbuffered (and to avoid having to write a fflush(stdout) after every output operation), use the ISO C setvbuf function declared in stdio.h:
setvbuf(stdout, (char *)NULL, _IONBF, 0);
once before doing the first I/O.
I was trying something when I saw this strange behavior. Can someone please explain it to me.
#include<stdio.h>
int main()
{
printf("utkarsh");
sleep(10);
printf("dixit");
}
The expected output is print "utkarsh" then wait for 10 seconds, print "dixit" next to it.
But what I observed was that it waits for 10 seconds and the prints "utkarshdixit".
If I add a \n at end of utkarsh, it works as expected.
printf("utkarsh\n");
Could someone help me understand why am I seeing such behavior ?
you're encoutering buffering.
try to do
fflush(stdout);
in front of the sleep
printf is buffering until you decide to write a '\n'.
You can use : flush to force to print
Here, try this
#include<stdio.h>
int main()
{
printf("utkarsh");
fflush(stdout);
sleep(10);
printf("dixit");
}
There's buffering of the standard out going on. you have to explicitly flush it.
There is buffering in stdout stream. Hence you need to flush once before sleep.
But when you used a '\n', the c run-time automatically flushes the stdout buffer for you. Hence you see this behaviour
I want to turn off the buffering for the stdout for getting the exact result for the following code
while(1) {
printf(".");
sleep(1);
}
The code printf bunch of '.' only when buffer gets filled.
You can use the setvbuf function:
setvbuf(stdout, NULL, _IONBF, 0);
Here're some other links to the function.
POSIX
C/C++
You can also use setbuf
setbuf(stdout, NULL);
This will take care of everything
Use fflush(FILE *stream) with stdout as the parameter.
http://www.elook.org/programming/c/fflush.html
You can do this:
write(1, ".", 1);
instead of this:
printf(".");
Use fflush(stdout). You can use it after every printf call to force the buffer to flush.
The following code:
int z = 0;
while(z < 4)
{
printf("iteration %d\n",z);
sleep(1);
z++;
}
Works fine and stdout buffer is flushed every second if running the program from command line. However, when I try to access the program in a web browser (server - apache on linux, compiled executable (with gcc) handled through cgi), the content is displayed only after 4 seconds and not "step by step". I am looking for something like PHP's ob_flush(). And, by the way, is cgi the best way of processing compiled C executables?
Update: neither fflush(stdout) nor setvbuf(stdout, NULL, _IONBF, 0) is working!!! Works great after disabling mod_deflate.
I am not quite sure I understand your question correctly, but in C you can
Flush after each print (fflush)
Disable buffering (setbuf, setvbuf)
setvbuf(stdout, NULL, _IONBF, 0); /* this will disable buffering for stdout */
If these won't work, then either something else is doing buffering or buffering is not the problem.
You could try to fflush stdout after your printf.