This question already has answers here:
What are the rules of automatic stdout buffer flushing in C?
(5 answers)
printf flush at program exit
(4 answers)
Closed 3 years ago.
Until now, I used fflush(stdout) when using printf without \n, since we have to flush the buffer.
But I found that the below code also works fine.
#include <stdio.h>
int main(void){
printf("hello world");
// without fflush(stdout)
}
// output: "hello world"
Is the outcome differs with the compiler? or am I misunderstand something?
Related
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 3 years ago.
In ubuntu,My code is:
int main(){
printf("signal test");
while(1);
return 0;
}
and it can'd display: signal test
This is because of buffering, as no newline is printed, nothing is flushed to the output.
This question already has answers here:
How to overwrite stdout in C
(8 answers)
Closed 4 years ago.
I know how to print string or chars in C, but i'm wondering how i could modify a string that have already been printed in the screen (like when you install some packages and # fills the |####---->| 50%), without using any other functions than syscalls.
You can use the carriage return:
#include <stdio.h>
#include <unistd.h>
int main()
{
printf("%s", "Hello, ");
fflush(stdout);
sleep(1);
printf("\r%s\n", "World!");
return 0;
}
This question already has answers here:
printf anomaly after "fork()"
(3 answers)
Closed 6 years ago.
i wanted to find someone that can explain this to me. I have this program:
int main(int argc, char *argv[]){
printf("P ");
if(fork()==0)
printf("C ");
return 0;
}
The result of this program is: P P C
What's the reason for that second "P" ?
IO buffering is the reason. printf is not printing the text right away, but waiting for newline, fflush or the end of the program to actually print it. But the buffer for the "future-to-print" text is in the memory that is getting duplicated by fork, so both processes receive it. And in the end both are printing it.
This question already has answers here:
printf not printing to screen
(3 answers)
Why does printf not flush after the call unless a newline is in the format string?
(10 answers)
Closed 7 years ago.
I have code like this below:
#include <stdio.h>
int main(int argc, char *argv[]) {
printf("hello");
while(1){
// whatever here
}
}
and the question is: why the first instruction is skipped? It runs only the loop, hello is never printed. I compiled it with gcc and g++ with the same result.
It does run, it's just that the output buffer is not flushed before your while.
Use printf("hello\n"); instead. The newline character will flush the buffer so writing the output immediately to your console.
Your assumption is wrong, your code does run, only stdout is not flushed, but buffered.
Use fflush(stdout) after printf("hello"), this forces stdout to be printed.
And, as #Bathsheba pointed out, also a newline character ("\n") within the printf forces it to flush, which is explained in this SO question.
This question already has answers here:
printf anomaly after "fork()"
(3 answers)
fork() in c using printf [duplicate]
(2 answers)
Closed 8 years ago.
#include <stdio.h>
#include <unistd.h>
int main(void)
{
int i;
for (i = 0; i < 2; i++) {
fork();
printf("-");
}
return 0;
}
The result of this program is 8"-" : "--------". But if I change 'printf("-");' into 'printf("-\n");', the result of this program will become 6"-": "-\n-\n-\n-\n-\n-\n". Can anyone tell me why?
printf writes to stdout stream which is line buffered. Buffer is a block of memory which belongs to a stream and is used to hold stream data temporarily. This is done to increase efficiency, as file and console I/O is slow in comparison to memory operations. Line buffered means that characters are saved up in the buffer only till a newline is output.