On many systems printf is buffered, i.e. when you call printf the output is placed in a buffer instead of being printed immediately. The buffer will be flushed (aka the output printed) when you print a newline \n.
On all systems, your program will print despite the missing \n as the buffer is flushed when your program ends.
Typically you would still add the \n like:
printf ("%s\n", a);
An alternative way to get the output immediately is to call fflush to flush the buffer. From the man page:
For output streams, fflush() forces a write of all user-space buffered data for the given output or update stream via the stream's underlying write function.
Source: http://man7.org/linux/man-pages/man3/fflush.3.html
EDIT
As pointed out by @Barmar and quoted by @Alter Mann it is required that the buffer is flushed when the program ends.
Quote from @Alter Mann:
If the main function returns to its original caller, or if the exit function is called, all open files are closed (hence all output streams are flushed) before program termination.
See calling main() in main() in c
On many systems printf is buffered, i.e. when you call printf the output is placed in a buffer instead of being printed immediately. The buffer will be flushed (aka the output printed) when you print a newline \n.
On all systems, your program will print despite the missing \n as the buffer is flushed when your program ends.
Typically you would still add the \n like:
printf ("%s\n", a);
An alternative way to get the output immediately is to call fflush to flush the buffer. From the man page:
For output streams, fflush() forces a write of all user-space buffered data for the given output or update stream via the stream's underlying write function.
Source: http://man7.org/linux/man-pages/man3/fflush.3.html
EDIT
As pointed out by @Barmar and quoted by @Alter Mann it is required that the buffer is flushed when the program ends.
Quote from @Alter Mann:
If the main function returns to its original caller, or if the exit function is called, all open files are closed (hence all output streams are flushed) before program termination.
See calling main() in main() in c
Strangely enough, it seems that nobody posted the adjusted code where the buffer is flushed yet...:
#include <stdio.h>
int main (void)
{
char a[]="abcde";
printf ("%s", a);
fflush(stdout);
//On some systems the line above will fail, in that case use: fflush(NULL);
}
Also note that this code probably doesn't do what you actually want to do.
What I assume you really want to do is:
#include <stdio.h>
int main (void)
{
char a[]="abcde";
printf ("%s\n", a);
//The '\n' makes sure the next thing you print will be on the following line
}
Hi guys,
I have an issue while trying to use the printf. For any reason is not showing no matter in my code.
Could you have a fast look and give me advice of why it's not working? Program simply exits with 0 exit code.
https://i.imgur.com/tQi2QLO.png
I've tried even using printf after the int main(){ but neither is printing anything after running the code through command line!!
speller- tried using printf to debug -printf not working - CS50 Stack Exchange
c - printf not printing to screen - Stack Overflow
C - printf() not working but puts() is working fine - Stack Overflow
c - printf not printing on console - Stack Overflow
Like @thejh said your stream seems to be buffered. Data is not yet written to the controlled sequence.
Instead of fiddling with the buffer setting you could call fflush after each write to profit from the buffer and still enforce the desired behavior/display explicitly.
printf( "Enter first integer\n" );
fflush( stdout );
scanf( "%d", &i1 );
you can try for disabling the buffering in stdout by using
setbuf(stdout, NULL);
Because printf() does not flush the output stream automatically. On the other hand puts() adds a new line '\n' at the end of the passed string. So it's working because the '\n' flushes de stdout.
Try
printf("hello\n");
Or, explicitly flush stdout
fflush(stdout);
right after the printf() statement.
Try using the new line character ('\n') at the end of your statement, also make sure you have the appropriate headers.
Output is buffered.
stdout is line-buffered by default, which means that '\n' is supposed to flush the buffer. Why is it not happening in your case? I don't know. I need more info about your application/environment.
However, you can control buffering with setvbuf():
setvbuf(stdout, NULL, _IOLBF, 0);
This will force stdout to be line-buffered.
setvbuf(stdout, NULL, _IONBF, 0);
This will force stdout to be unbuffered, so you won't need to use fflush(). Note that it will severely affect application performance if you have lots of prints.
Apparently this is a known bug of Eclipse. This bug is resolved with the resolution of WONT-FIX. I have no idea why though. here is the link: Eclipse C Console Bug.
code
# include<stdio.h>
int main()
{
int size;
printf("Enter the number of elements\n");
scanf("%d",&size);
int a[size];
printf("Enter the elements\n");
for(int i = 0 ;i<size;i++)
{
scanf("%d",&a[i]);
}
printf("Before sorting\n");
for(int i = 0 ;i<size;i++)
{
printf("%d\n",a[i]);
}
for(int i = 0 ;i<size;i++)
{
printf("%d\n",a[i]);
}
return 0;
int min;
for(int i=0;i<size;i++)
{
for(int j=i;j<size;j++)
{
if(a[i]<a[j])
{
min = a[i];
}
else
{
min = a[j];
}
}
a[i]=min;
}
printf("After sorting\n"); // This is't working
for(int i = 0 ;i<size;i++)
{
printf("%d\n",a[i]);
}
return 0;
}
Output
Enter the number of elements
5
Enter the elements
23
432
535
234
552
Before sorting
23
32
35
234
552
23
32
35
234
552
Thank you for your time
EDIT: As most are pointing out, it probably has to do with non-printable characters (eg, going above ASCII value 122) because I'm remembering that at on point it was able to print out the character "h" (number 104) so I'll work on that soon.
Hi, this is probably a very beginner question but it´s all about C so I didn´t know if to post in r/learnprogramming. The answers on Google seem to be for similar but more complicated issues.
Some of you might recognize this as a CS50 exercise from week 2 but essentially I cannot get the integer encryptedtext[i] to be recognized and printed out as a character. Below is my code, it is an early draft so right now I´m only looking for help around (I think) line 36:
#include <stdio.h>
#include <cs50.h>
#include <string.h>
#include <stdlib.h>
int main(int argc, string argv[])
{
// user may only input 2 command-line arguments
if(argc != 2)
{
printf("Restart program and input one key only. \n");
return 1;
}
string key = argv[1];
int keylength = strlen(key);
for(int i = 0; i < keylength; i++)
{
if(key[i] < 65 || (key[i] > 90 && key[i] < 97) || key[i] > 122)
{
printf("Restart program and input letters only. \n");
return 1;
}
}
printf("Give me something to encrypt: ");
string plaintext = GetString();
//looping through key and plaintext
int plaintextlength = strlen(plaintext);
int encryptedtext[plaintextlength];
for(int i = 0; i < plaintextlength; i++)
{
encryptedtext[i] = plaintext[i]+key[i];
//printf("Your plaintext %c added to key %c becomes: %c \n", plaintext[i], key[i], encryptedtext[i]);
printf("%c is %i. \n", encryptedtext[i], encryptedtext[i]);
}
return 0;
}As far as I know, encryptedtext[i], plaintext[i], and key[i] are all integers, or can be treated as such. In fact, it prints the ASCII number value of the letter but not the letter itself. It probably has to do with me not understanding exactly how characters and integers can be used interchangeable (edit: actually it probably has to do with unprintable ASCII values) but I've tried declaring encryptedtext[] as an array of characters and I've played around with different references (%c, %s, %i, %d) but I can´t get it to print out the character! Any help would be appreciated.
Thanks!