This question already has answers here:
Why does printf not flush after the call unless a newline is in the format string?
(10 answers)
Closed 7 years ago.
pid_t pid=fork();
printf("%d\n",pid);
if(pid==0){
sleep(3);
printf("!");
}
else
{
printf("#");
read_routine(clnt_sock,buf);
}
On my console, I can see two pid and !, but there is no #.
And when I delete the statement read_routine(clnt_sock,buf);, then I can see # on console.
In read_routine function, there is just some input statement using fgets().
Are there some secrets of printf?
Your output is probably buffered. After the printf, you will likely want to fflush(stdout);.
Add fflush(stdout) after print and try.
Related
This question already has an answer here:
C/Unix Strange behaviour while using system calls and printf
(1 answer)
Closed 3 years ago.
I just started using the system() function in c, and I thought of starting the same executable from within it self using the system function, so I wrote the following program
#include <stdlib.h>
#include <stdio.h>
int main()
{
printf("some string");
system("./a.out");
}
-I used gcc to compile it-
when I ran the program it did not print anything, it just kept going until I used the shortcut ctrl-c to stop the execution,then it started printing the output(it did not print anything until I stopped it)
I believe the statements should execute sequentially, why did it not print anything until I stopped it?
By default, when stdoutis connected to a terminal, it is line-buffered.
printf("some string");
doesn't have a '\n' in it and you aren't calling fflush(stdout); after it either, so all this printf("some string"); does is copy "some string" into your stdout's output buffer.
The buffer is flushed as the end of main.
printf("some string\n"); would flush the buffer immediately, provided stdout is connected to a terminal and you didn't change stdout's buffering.
printf("some string"); fflush(stdout); will flush the buffer immediately regardless of context and without the need for the '\n'.
This question already has an answer here:
Why no output on console on signal handling?
(1 answer)
Closed 4 years ago.
I want to test if different loops are active, so I have a print statement that repeats every 500ms in each loop, however the print statement does not print every 500ms, it waits until the loop is finished and then prints everything at once instead of periodically.
How do I get my program to print to the terminal periodically?
I'm a student so my knowledge of SDL is pretty limited so thorough explanation would be appreciated.
int main(void)
{
int i = 0;
while(i<10)
{
printf("While loop active.\t"); i++;
SDL_Delay(500);
}
return 0;
}
P.S. I saw that duplicate question but I disagree as his question suggests an issue with 'signal handling' which I know nothing about so when asking this question, I didn't think his question would have the same answer as mine. I accept the answers given are the same though.
You are probably not flushing stdout. Add a newline to the printf call and you should be OK:
printf("While loop active.\n");
/* Here ------------------^ */
Or if keeping \t is essential:
printf("While loop active.\t");
fflush(stdout);
Credit to #lurker for extra information.
This question already has answers here:
printf just before a delay doesn't work in C
(5 answers)
Closed 5 years ago.
Why below function is not printing "Just kidding!".
void justCheck() {
printf("Just kidding!");
while (1) {
}
}
while this is printing "Justing Kidding!" followed by non stopping "Just Kidding inside loop!".
void justCheck() {
printf("Just kidding!\n");
while (1) {
printf("Justing Kidding inside loop!\n");
}
}
can anyone please explain the logic?
Your first example
printf("Just kidding!");
The output is buffered and therefore not displayed
In the second example
printf("Just kidding!\n");
The \n at the end will flush the buffer and therefore the string will be displayed.
In the first example before the while loop insert fflush(stdout);
This question already has answers here:
Why does printf not flush after the call unless a newline is in the format string?
(10 answers)
Closed 6 years ago.
I can't understand why following code works like this..I mean: instead of printing "hello" after each one second delay...it wait 5 second and display hellohellohellohellohello at once.
#include <stdio.h>
int i;
for(i=0; i<5; i++) {
printf("hello");
sleep(1);
}
The output of printf() (stdout) is line buffered by default if the output is going to a tty. You need one of
printf("hello\n");
or
printf("hello");
fflush(stdout);
The latter will explicitly flush the output each iteration.
printf do not print immediatly, instead it cache it line per line.
Add "\n" (newline) add the end of your string printf("hello\n"); or use write function instead write(STDOUT_FILENO, "hello", sizeof("hello"));.
You are writing to standard output (stdout), which is buffered. If you want the content to be immediately printed, you may flush the output or insert a newline.
You can add a \n to the end of your string so as to print the newline -- change your printf line to:
printf("hello\n");
For flushing the stdout buffer call fflush on it, after the printf:
#include <stdio.h>
int main() {
int i;
for(i=0; i<5; i++) {
printf("hello");
fflush(stdout);
sleep(1);
}
}
In general the output can be buffered. That means the implementation collects several bytes before actually writing to the console. You can explicitly write the buffer by fflush(stdout). This is true for all file descriptors, one of which is stdout, the terminal output. You can disable the buffering with setbuff(stdout, NULL), but this is almost never a good idea performance wise.
Try this:
int i;
for(i=0;i<5;i++){
printf("hello\n");
i=0;
sleep(1);
}
This question already has answers here:
Why does printf not flush after the call unless a newline is in the format string?
(10 answers)
Closed 8 years ago.
#include <stdlib.h>
int main()
{
printf("\nHello");
sleep(5);
printf("\nLinux");
}
In my expectation, It should be like:
PRINT Hello --- WAIT 5 SECS ---> PRINT Linux
But actually it will be like this:
WAIT 5 SECS --> PRINT Hello --> PRINT Linux
Why ? How to make my program be the first one (as my expectation) ?
And why my code can run expectedly on Win32 Console?
Your stream is line-buffered, as you don't end your string with \n, flush it using fflush.
Change your program to:
int main()
{
printf("\nHello");
fflush(stdout);
sleep(5);
printf("\nLinux");
}
The output is buffered and is not printed until newline.
Try with:
printf("\nHello\n");
sleep(5);
printf("Linux");