Hi i am new to the programing and for example in my code:
#include <stdio.h>
int main (void){
int a;
printf("Write a number: ");
scanf("%d", &a);
printf("Your written number was: %d", a);
return 0;
}
Printf does not write "write a number" in console when i start the program but only after i already inserted the number and pressed enter.
I have already done some research and found out for this code:
setvbuf(stdout, NULL, _IONBF, 0);
when i paste this into my program it works as it should but i am wondering why do i have to do that?
when i paste this into my program it works as it should but i am
wondering why do i have to do that?
It's because printf() is usually line-buffered when attached to a terminal. So disabling the buffering with the call to setvbuf() makes stdio library to not buffer at all.
You can also use fflush(stdout); after the printf() call to flush out the buffered output. The same can be done with setbuf(stdout, NULL); as well.
You can also add a \n at the end of printf() statement to force the flushing. But this will work only if the output goes to a terminal device.
For example, if you do (on a unix-like system):
./a.out > output_file
then the \n will not flush the buffer.
Out of the two options (setbuf() and fflush()),fflush(stdout); is probably the better option in most cases. Since disabling the buffering completely can have negative impact on performance (which is the primary reason for buffering in the first place) whereas fflush() can be judiciously used at the right place when you think it's necessary.
printf has a buffer. It is a mechanism to make code run faster by not having to switch between the user context and the kernel context. To get over this you can tell the code to flush the buffer - i.e. send it to the operating system. This can be done by
fflush(stdout);
After a printf. If the printf contains a new line this is done automatically.
You probably want \n in each of those printf statements.
Add linefeed "\n" to your printf lines like so:
printf("Write a number: \n");
Related
#include <stdio.h>
#define MAXLEN 256
int main() {
int n;
char buf[MAXLEN];
while((n = read(0,buf,sizeof(buf))) != 0){
printf("n: %d:",n);
write(1,buf,n);
}
return 1;
}
The output of the program (where the first read and first write is typed by the user and echoed by the terminal) is:
read
read
write
write
n: 5:n: 6:
The output of printf comes after pressing Ctrl+D at the standard input and not along with the subsequent reads. Why does this happen?
Printf is buffered.
You can force printf to 'flush' its buffer using the fflush call:
#include <stdio.h>
#define MAXLEN 256
int main() {
int n;
char buf[MAXLEN];
while((n = read(0,buf,sizeof(buf))) != 0){
printf("n: %d:",n);
fflush(stdout); /* force it to go out */
write(1,buf,n);
}
return 1;
}
In general, printf() being buffered is a good thing. Unbuffered output, particularly to visible consoles that require screen updates and such, is slow. Slow enough that an application that is printf'ing a lot can be directly slowed down by it (especially on the Windows platform; Linux and unixes are typically impacted less).
However, printf() being buffered does bite you if you also fprintf(stderr,) - stderr is deliberately unbuffered. As a consequence, you may get your messages with some printf() missing; if you write to another FILE handle that is also associated with the terminal, and might be unbuffered, make sure you first explicitly fflush(stdout).
The manpage for fgets tells me:
It is not advisable to mix calls to input functions from the stdio
library with low-level calls to read(2) for the file descriptor associ‐
ated with the input stream; the results will be undefined and very
probably not what you want.
So the best solution would be not to to use write and printf on the same descriptor.
Printf is using stdio and it is buffered.
Push it out by sending a changing to "n: %d:\n"
You can use the std fflush() function to flush the std out buffer or you can make use of an additional \n at the end of the control string inside the printf. Something like this
printf("\n :%d:\n",n);
Its always better to use the write() & read() functions in C instead of printf() and scanf(). Printf and scanf have got some problems like printf stores the string parameter in stdout buffer. So a manual flush is required which is done through fflush function or by means of \n. In a small hello world printing program you will not find such a problem as the stdout buffer is flushed at the end of the program execution. Better use write() which works fine. scanf also have the problem of reading spaces and a lot of other problems related to stdin buffer.
For example in the code below:
main() { char a; int i=0,c; for(;i<2;i++) { scanf("%d",&c); scanf("%c",&a);} }
The above program as got the problem of reading \n into stdin on pressing enter. We could resolve this but not flushing the stdin buffer or making use of \n character. Always better to use the read() and write() functions.
Hope that helps....
Use fwrite (streams version) rather than write.
Note that, while is associated with file number 1, it isn't the same thing.
I'm running Netbeans 8.1 on a Windows 10 x86 64-bit computer, I have the project's Run>Console Type set to Standard Output because neither internal nor external terminal have worked for me with any other source I've made--for which standard output had. In any case, here is the code I'm attempting to run:
int main(void){
float original_amount, amount_with_tax;
printf("Enter an amount: ");
scanf("%f", &original_amount);
amount_with_tax = original_amount * 1.05f;
printf("With tax added: $%.2f\n", amount_with_tax);
exit(EXIT_SUCCESS);
}
and here is the output:
3
Enter an amount: With tax added: $3.15
RUN SUCCESSFUL (total time: 4s)
As you can see, the scan function is reading in the number before the program even prints "Enter an amount:". Also, after I commented out the scanf function, it printed both printf statements as expected. I have been wrestling with this problem for a while now and any help is appreciated, thanks!
ISO/IEC 9899:201x:
Files
… When a stream is line buffered, characters are intended to be transmitted to or from the host environment as a block when a new-line
character is encountered. …
Your standard output stream seems to be line buffered, which is most often so.
try printf("Enter an amount: ");fflush(stdout); – BLUEPIXY
That worked! Is this only necessary because I'm using the standard output as my run console?
No, other streams opened by your program may even be fully buffered.
And if so, should I just put it at the beginning of any program that
will use a function accessing the buffer?
Putting fflush(stdout) at the beginning of a program won't do, since it outputs only once what is already in the stdout buffer. But you can use setbuf(stdout, NULL) there.
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
#include <stdio.h>
#define MAXLEN 256
int main() {
int n;
char buf[MAXLEN];
while((n = read(0,buf,sizeof(buf))) != 0){
printf("n: %d:",n);
write(1,buf,n);
}
return 1;
}
The output of the program (where the first read and first write is typed by the user and echoed by the terminal) is:
read
read
write
write
n: 5:n: 6:
The output of printf comes after pressing Ctrl+D at the standard input and not along with the subsequent reads. Why does this happen?
Printf is buffered.
You can force printf to 'flush' its buffer using the fflush call:
#include <stdio.h>
#define MAXLEN 256
int main() {
int n;
char buf[MAXLEN];
while((n = read(0,buf,sizeof(buf))) != 0){
printf("n: %d:",n);
fflush(stdout); /* force it to go out */
write(1,buf,n);
}
return 1;
}
In general, printf() being buffered is a good thing. Unbuffered output, particularly to visible consoles that require screen updates and such, is slow. Slow enough that an application that is printf'ing a lot can be directly slowed down by it (especially on the Windows platform; Linux and unixes are typically impacted less).
However, printf() being buffered does bite you if you also fprintf(stderr,) - stderr is deliberately unbuffered. As a consequence, you may get your messages with some printf() missing; if you write to another FILE handle that is also associated with the terminal, and might be unbuffered, make sure you first explicitly fflush(stdout).
The manpage for fgets tells me:
It is not advisable to mix calls to input functions from the stdio
library with low-level calls to read(2) for the file descriptor associ‐
ated with the input stream; the results will be undefined and very
probably not what you want.
So the best solution would be not to to use write and printf on the same descriptor.
Printf is using stdio and it is buffered.
Push it out by sending a changing to "n: %d:\n"
You can use the std fflush() function to flush the std out buffer or you can make use of an additional \n at the end of the control string inside the printf. Something like this
printf("\n :%d:\n",n);
Its always better to use the write() & read() functions in C instead of printf() and scanf(). Printf and scanf have got some problems like printf stores the string parameter in stdout buffer. So a manual flush is required which is done through fflush function or by means of \n. In a small hello world printing program you will not find such a problem as the stdout buffer is flushed at the end of the program execution. Better use write() which works fine. scanf also have the problem of reading spaces and a lot of other problems related to stdin buffer.
For example in the code below:
main() { char a; int i=0,c; for(;i<2;i++) { scanf("%d",&c); scanf("%c",&a);} }
The above program as got the problem of reading \n into stdin on pressing enter. We could resolve this but not flushing the stdin buffer or making use of \n character. Always better to use the read() and write() functions.
Hope that helps....
Use fwrite (streams version) rather than write.
Note that, while is associated with file number 1, it isn't the same thing.
I'm just learning C with Kernighan and Ritchie's book; I'm in the basics of the fourth chapter ("Functions and Program Structure"). The other day I became curious about the sleep() function, so tried to use it like this:
#include <stdio.h>
#include <unistd.h>
int main(void)
{
printf(" I like cows.");
sleep(5);
return 0;
}
The problem is the output of the program, it looks like it does the sleep() first and then the printf(), in other words, it waits five seconds and then prints the string. So I thought, maybe the program gets to sleep() so fast that it doesn't let printf() have his work done like I want, that is print the string and then sleep.
How can I show the string and then put the program to sleep?
The compiler is GCC 3.3.5 (propolice) in OpenBSD 4.3.
printf() writes to stdout (the default output stream) which is usually line buffered. The buffer isn't flushed by the time sleep is called so nothing is displayed, when the program exits all streams are automatically flushed which is why it prints right before exiting. Printing a newline will usually cause the stream to be flushed, alternatively you could use the fflush function:
int main(void)
{
printf(" I like cows.\n");
sleep(5);
return 0;
}
or:
int main(void)
{
printf(" I like cows.");
fflush(stdout);
sleep(5);
return 0;
}
If you are printing to a stream that is not line buffered, as may be the case if stdout is redirected or you are writing to a file, simply printing a newline probably won't work. In such cases you should use fflush if you want the data written immediately.
Your problem is that printf (and anything else that uses the stdio library to write to stdout (standard output)) is buffered - line buffered if it goes to the console, and size buffered if it goes to a file. If you do a fflush(stdout); after the printf, it will do what you want. You could try just adding a newline ('\n') to your string, and that would do the right thing as long as you don't redirect standard output to a file.
I'm not 100% sure, but I think stderr isn't buffered, which can cause confusion because you might see output you made to stderr before output you previously made to stdout.
Buffering means that all the output is stored in a place (called buffer) and is output after a certain amount of data is present in it. This is done for efficiency reasons.
Some (most?) implementations clear the buffer after a newline when writing to the console, so you can also try
printf(" I like cows.\n");
instead of the call to fflush()
I implemented time encounter as following;
for (int i = 1; i <= 60; i++) {
printf("%02d", i);
fflush(stdout);
sleep(1);
printf("\b\b");
}