While Trying this code on eclipse CDT with GCC 5.1.0 Compiler
All the strings were printed after the user input ..
and while compiling it on Visual Studio and Code Blocks IDEs even with the windows CMD The program worked just fine as expected ..
#include <stdio.h>
static char string[128] = "";
int main() {
printf("Type a string: ");
scanf("%s",string);
printf("The String is %s", string);
return 0;
}
Eclipse Output:
Visual Studio Output:
Thanks ,,,
OK, I see now. I think the issue is that whenever you want to be certain that something is printed by a given point in the code, you need to flush stdout at that point.
Otherwise, streamed contents can be queued and delivered in an implementation-dependent way (usually in small batches)
The C standard library's printf(), when outputting to stdout and encountering a newline \n, provides an implicit flush, so you don't need to call flush() yourself. Whereas with C++'s std::cout, only std::endl has this property; \n is not guaranteed to.
Deliberately flushing stdout in C can be done like so: fflush(stdout);
See also: Why does printf not flush after the call unless a newline is in the format string?
Related
#include <stdio.h>
int main(){
char a[2] = {0};
a[0] = 't';
printf("%s", a);
scanf("%c", a);
return 0;
}
scanf here will cause an automatic flush of stdout.
Running a.out will print t on the terminal before running scanf, more info: How is printf getting flushed before scanf is executed?
However, doing a.out > out.txt and terminating it with ^C does not print anything inside out.txt yet the output still appeared on the screen without redirecting stdout with >.
If stdout is being flushed then why out.txt is still empty?
If it's not being flushed then how did t appear on the screen in the first example?
(I know using \n or a manual fflush or properly terminating the program will fix the issue, i'm just curious about this behaviour).
The key is the word interactive:
The input and output dynamics of interactive devices shall take place as specified in 7.21.3.
As soon as you redirect the standard output to a file, it is no longer interactive.
For example the Linux C standard library actually executes the analogue of the isatty library call to figure this out. If it figures out that standard output is not directed to file it will also break this relationship. It will also increase the I/O performance of programs that work as part of a command pipeline.
You can yourself test whether stdout is connected to a terminal by executing
#include <unistd.h>
printf("stdout is connected to a terminal: %d\n", isatty(fileno(stdout)));
This question already has answers here:
Why doesn't getchar() wait for me to press enter after scanf()?
(10 answers)
Closed 3 years ago.
I'm learning programmation in C and tried to create a program that asks the user his age. When the user writes his age (for example 18) he gets the message "So you're 18 years old". When I execute the .exe file it automatically closes after you see the message, so fast that you don't see it. Then I added getchar so that the user reads the message and then presses Enter to quite. Here's the program I wrote:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int age=0;
printf("How old are you?\n");
scanf("%d",&age);
printf("So you're %d years old", age);
getchar();
return 0;
}
Unfortunately, when I execute the .exe file, it still closes automatically like if the getchar() doesn't exist and I don't know why.
scanf("%d",&age);
When the execution of the program reaches the above line,you type an integer and press enter.
The integer is taken up by scanf and the \n( newline character or Enter )which you have pressed remains in the stdin which is taken up by the getchar().To get rid of it,replace your scanf with
scanf("%d%*c",&age);
The %*c tells scanf to scan a character and then discard it.In your case,%*c reads the newline character and discards it.
Another way would be to flush the stdin by using the following after the scanf in your code:
while ( (c = getchar()) != '\n' && c != EOF );
Note that c is an int in the above line
You're only having trouble seeing the result because you're starting the program from a windowing environment, and the window closes as soon as its internal tasks are completed. If you run the compiled program from a command line in a pre-existing shell window (Linux, Mac, or Windows), the results will stay on the screen after you're returned to the prompt (unless you've ended by executing a clear-screen of some sort). Even better, in that case, you don't need the extraneous getchar() call.
For Windows, after opening the command prompt window, you'd issue a "cd" command to change to the directory that contains the compiled program, and then type its name. For Linux (and, I presume, Mac, since Mac is UNIX under the hood), you'd need to type ./ ahead of the program name after changing to the appropriate directory with "cd".
This question already has answers here:
Simple program adding "D" to output
(3 answers)
Why does C program print 0D instead of 0? (When EOF sent as Ctrl+D) [duplicate]
(1 answer)
Closed 5 years ago.
I was coding a very simple programme to detect word pattern by entering to stdin and return the times found the pattern.
However the code return me the correct number but follow a char D.
#include <stdio.h>
#include "string.h"
#define MAXLINE 1000 /* maximum input line length */
char pattern[] = "ould"; /* pattern to search for */
/* print all lines from standard input that match pattern */
int main()
{
char line[MAXLINE];
int found = 0;
while (fgets(line, MAXLINE, stdin) != NULL)
if (strstr(line, pattern) != NULL) {
printf("%s", line);
found++;
}
printf("%d \n", found);
return 0;
}
Result:
glaroam2-180-76:Lab2 apple$ ./find0
fould
fould
1D
The code is correct (apart from the #include "string.h" which should be
#include <string.h>)1, the problem is that when you press
Ctrl+D on your terminal, your terminal might write
something on the terminal, which you cannot control and this output might be
^D
After fgets returns NULL, you do printf("%d \n", found); which prints the '1'.
But because there was ^D on the terminal, the ^ was replaced by the '1' and
you end up with:
1D
Change your last printf to this:
printf("\n\n%d \n, found);
And you might see only a '1' in the next lines of the output.
This has nothing to do with your C program, it's the behaviour of your terminal.
My terminal for example doesn't print when pressing Ctrl+D,
but when pressing Ctrl+C I get ^C. There's nothing you
can do.
edit
With There's nothing you can do I mean that you cannot control the way the
terminal from you C program without calling external tools like stty. While
this might solve your problem, you are loosing portability.
However, before you start you program, you can configure your terminal using a
program like stty. See Jonathan Leffler's answer for more info on that.
Fotenotes
1As Jonathan Leffler points out in the comments, using quotes instead
of angle brackets for system headers is not an error per se. For example my GCC
compiler searches in the same directory of the compiled file for headers that
were included with quotes. But in general, it's a good practice to include the
header files included provided by your system with angle brackets.
It's a terminal setting: echoctl. It means that when you type Control-D, the terminal echoes ^D, and then the 1 overwrites the ^. Try using:
stty -echoctl
and then rerunning your program.
With that said, I'm surprised that the D isn't wiped out by the blank after the %d in the format string. I suspect your actual code may be missing that. When I tested on my Mac, the program with the space after the %d did not show the D for long enough for me to spot it; when I removed that space, I got the output shown in the question.
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'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");
}