Problems with C scanf("%c") function to read characters one by one
The following code produces a very strange result when I run it.
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
for ( ; ; )
{
char test;
printf("Please enter 'w' ");
scanf("%c", &test);
printf("%c\n", test);
if (test == 'w')
{
printf("Working\n");
}
else
{
printf("ERROR\n");
return 0;
}
}
}
What I want to happen is for the whenever I input 'w' it continues the loop so I can input 'w' again. What it does though is go to the else statement even though I input 'w'. It just seems to skip the scanf()
line. I have asked everyone I know who knows C but they do not know how to solve it.
Somebody please help me out here!
This is because you type w followed by ENTER. So there are 2 characters in the input, 'w'
, followed by a newline (\n
). The latter causes the else
branch to be taken on the second iteration.
Note that standard input is line buffered when connected to a terminal. If you need to deal with characters immediately, there are ways to do that. See the comp.lang.c FAQ for details ("How can I read a single character from the keyboard without waiting for the RETURN key? How can I stop characters from being echoed on the screen as they're typed?").
Note that for robust programming it is a must to check the return value of scanf
. It returns the number of successfully converted items. As shown, your code does not handle the case of end-of-file properly, i.e. when the user types Ctrl-D (assuming Unix terminal). Then scanf returns EOF
and no conversion was performed, but you use test
as if it contained a meaningful value.
as Jens said. you have to ignore the newline '\n'
Adding a space at the beginning of the format specifier " %c"
will ignore the newline '\n'
scanf(" %c", &test);
Using " %c"
will also ignore other white spaces like \t
space
\b
\v
\r