Problem With Using fgets()/gets()/scanf() After scanf() in C
Last Updated :
04 Jan, 2022
scanf() is a library function in C. It reads standard input from stdin. fgets() is a library function in C. It reads a line from the specified stream and stores it into the string pointed to by the string variable. It only terminates when either:
- end-of-file is reached
- n-1 characters are read
- the newline character is read
1) Consider a below simple program in C. The program reads an integer using scanf(), then reads a string using fgets(),
Input
10
test
C
#include <stdio.h>
int main()
{
int x;
char str[100];
scanf ( "%d" , &x);
fgets (str, 100, stdin);
printf ( "x = %d, str = %s" , x, str);
return 0;
}
|
Output
x = 10, str =
Explanation: The problem with the above code is scanf() reads an integer and leaves a newline character in the buffer. So fgets() only reads newline and the string “test” is ignored by the program.
2) The similar problem occurs when scanf() is used in a loop.
Input:
a
b
q
C
#include <stdio.h>
int main()
{
char c;
printf ( "Press q to quit\n" );
do {
printf ( "Enter a character\n" );
scanf ( "%c" , &c);
printf ( "%c\n" , c);
} while (c != 'q' );
return 0;
}
|
Output
Press q to quit
Enter a character
a
Enter a character
Enter a character
b
Enter a character
Enter a character
q
Explanation: We can notice that the above program prints an extra “Enter a character” followed by an extra newline. This happens because every scanf() leaves a newline character in a buffer that is read by the next scanf.
How to Solve the Above Problem?
- We can make scanf() to read a new line by using an extra \n, i.e., scanf(“%d\n”, &x) . In fact scanf(“%d “, &x) also works (Note the extra space).
- We can add a getchar() after scanf() to read an extra newline.
The corrected programs for the above points will be,
1) scanf() when there is fgets() after it:
Input:
10
test
C
#include <stdio.h>
int main()
{
int x;
char str[100];
scanf ( "%d\n" , &x);
fgets (str, 100, stdin);
printf ( "x = %d, str = %s" , x, str);
return 0;
}
|
Output
x = 10, str = test
2) When scanf() is used in a loop:
Input:
a
b
q
C
#include <stdio.h>
int main()
{
char c;
printf ( "Press q to quit\n" );
do {
printf ( "Enter a character\n" );
scanf ( "%c\n" , &c);
printf ( "%c\n" , c);
} while (c != 'q' );
return 0;
}
|
Output: Press q to quit
Enter a character
a
Enter a character
b
Enter a character
q
Must Read: Problem occurs with Scanner in Java when nextLine() is used after nextXXX()
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