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);
Related
This question already has answers here:
in c system() executes before printf() even when printf come first [duplicate]
(1 answer)
Why does printf not flush after the call unless a newline is in the format string?
(10 answers)
Closed 1 year ago.
the problem is when i run this code the system function start first and as per my understanding (which is very little), printf function should start first then the system function !
#include<stdio.h>
#include<stdlib.h>
int main()
{
printf("test1");
system("echo 'test2'");
}
and it output :
test2
test1
but the output should be like this instead!
test1
test2
You have to add a line break (\n) to the printf statement. I can only give you my basic (probably not entirely correct) explanation. printf gives the characters to a buffer for printing to the console. This buffer is then printed per line. If you add the \n the line is printed instantly.
#include<stdio.h>
#include<stdlib.h>
int main()
{
printf("test1\n");
system("echo 'test2'");
}
i just find the explination its when we say printf is actually what we call buffered the output that we ask to print to standard out in this case our display doesn't come to the display immediately it goes into a holding pool a buffer and when that buffer gets filled up it'll dump everything out at once it's an efficiency consideration but if we print out a new line that \n it will go ahead and it will flush the buffer out right ,so to fix it we can use :
#include<stdio.h>
#include<stdlib.h>
int main()
{
printf("test1\n");
system("echo 'test2'");
}
or :
#include<stdio.h>
#include<stdlib.h>
int main()
{
printf("test1");
fflush(stdout);
system("echo 'test2'");
}
both do the same job (flush the output buffer).
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:
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.
This question already has an answer here:
need of fseek() in c
(1 answer)
Closed 8 years ago.
while(1)
{
ch=fgetc(ft);
if(ch==EOF)
{
break;
}
if(ch=='u')
{
fputc('b',ft);
fflush(ft);
}
}
I tried to replace character after u with b in a file pointed by *ft.
This code runs fine but when I open the file it seemed to be unedited.
The above code works fine with fseeks(ft,0,SEEK_CUR).
Why it is not working with fflush(ft).
fflush only flushes output stream. Hence you need to put fseek(ft,0,SEEK_CUR) above your fputs(ft)
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");