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).
Related
This question already has answers here:
Why does printf() not print anything before sleep()?
(4 answers)
Closed 1 year ago.
I wrote a piece of code like this:
#include <stdio.h>
#include <unistd.h>
int main() {
printf("one");
sleep(1);
printf("\r");
printf("two");
return 0;
}
It was supposed to first print "one", and then override it by "two" after one second's pause.
But what really happened is that it only print "two" after one second's empty, without printing "one" first. I am confused, can someone figure it out? Thanks a lot.
The statement printf("one") does not immediately print "one" to the terminal, but it writes the string to the standard output stream (stdout) - which is usually buffered. Only when this stream is flushed, do its contents get printed to the terminal. Usually, this happens whenever a newline is printed or when the program exits, among other conditions.
The output stream can be forced to flush by calling fflush(stdout), which produces the desired result:
#include <stdio.h>
#include <unistd.h>
int main(void) {
printf("one");
fflush(stdout); // Line added
sleep(1);
printf("\r");
printf("two");
return 0;
}
See fflush for more information. The buffering mode of a stream can be set using the setvbuf function.
Also, note that the usual main signature, when using no parameters, is int main(void).
Acknowledgements: Thanks to Andreas Wenzel for a correction and the setvbuf addition.
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 1 year ago.
I want to print a line and then get it erased after ten seconds.
So I thought that I should write a timer as a preliminary programme and with erasure I could deal later using whatever I find on the internet (using \r for example).
And so I want to print "I love nachos" and ten seconds later "I hate nachos".
However, even though printf("I love nachos") is written before any conditions, the programme prints both strings after ten seconds.
#include <stdio.h>
#include <time.h>
int main(void) {
printf("I love nachos");
time_t now = time(NULL);
while (1)
{
time_t future = time(NULL);
if ((future-now)==10)
{
break;
}
}
printf("I hate nachos");
return 0;
}
I asked my friend to tackle the problem and he wrote:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main()
{
printf("Start\n");
time_t t0 = time(NULL);
while ( (difftime(time(NULL), t0)) <= 10.0 )
{}
printf("End");
return 0;
}
However, in that case the programme only works with \n in printf("Start\n").
If I remove \n I get the same result as per my initial code and I obviously don't want \n there, because I wouldn't be able to erase the line with \r
The printf output to stdout is by default line buffered. That means nothing is actually sent out until either the internal buffer is full or a newline is added or the FILE is closed (as happens implicitly when the program exits)
So either add a newline or use fflush to force the buffer to be transmitted immediately to the output.
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 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);
}
I know it is a simple question but I am stuck.The code is:
#include <stdio.h>
#define MAX_SIZE 1025
#define NUM 64
int main(){
int mem_size;
char types[NUM];
char values[MAX_SIZE];
fgets(types,NUM,stdin);
printf("%s",types);
fgets(values,MAX_SIZE,stdin);
printf("%s",values);
scanf("%d",&mem_size);
printf("%d",mem_size);
return 0;
}
Although I want the results after I type and hit enter, the flow is: I need to enter all the fgets and scanf stuff and it correctly prints the desired results.
What is the problem? Please help.
OP: "Problem is it shows the results after I enter mem_size altogether simultaneously, not one by one"
Some systems to not "flush" the stdout output promptly even with a \n. The output that was seen came out just before the program ended, which forced the buffered stdout to the console.
Either add fflush(stdout) after each printf() or change your system's settings (varies with environment) to promptly send stdout to the console.
Ref:
printf not printing on console