scanf asking for two values instead of one [duplicate] - c

Question

This question already has answers here:
scanf() curious behaviour!
(3 answers)
Closed 8 years ago.
When I compile this code, it leads to scanf asking for a value twice when I pick choice A. What am I missing here?
This isn't the first time this is encountered, so I'm suspecting I'm failing to grasp something rather fundamental with scanf.
#include <stdio.h>
#include <stdlib.h>
int main()
{
char choice;
printf("1. 'enter 'A' for this choice\n");
printf("2. 'enter 'B' for this choice\n");
printf("3. 'enter 'C' for this choice\n");
scanf("%c", &choice);
switch (choice)
{
case 'A':
{
int a =0;
printf("you've chosen menu A\n");
printf("enter a number\n");
scanf("%d\n", &a);
printf("%d\n", a);
}
break;
case 'B':
printf("you've chosen B\n");
break;
case 'C':
printf("you've chosen C\n");
break;
default:
printf("your choice is invalid\n!");
break;
}
return 0;
}

Answer 1

scanf("%d\n", &a); should be scanf("%d", &a);
Also read Related question.
In former case after reading an integer and storing into a, scanf's argument string is not exhausted. Looking at \n, scanf would consume all the whitespaces (newline, tab, spaced etc) it sees (And will remain blocked) until it encounters a non-whitespace character. On encountering a non-whitespace character scanf would return.
Learning: Don't use space, newline etc as trailing character in scanf. If the space character is in the beginning of argument string, scanf may still skip any number of white space characters including zero characters. But when whitespace is a trailing characters, it would eat your new line character also unless you type a non-whitespace character and hit return key.

Answer 2

Simply remove the newline character from scanf("%d\n", &a); and it will not ask to enter a value twice

Answer 3

scanf("%d", &a);
Remove a newline character while scanning

Answer 4

// the trailing '\n' in the scanf format string
// is what is causing the dual inputs
// the returned value from I/O statements (I.E. scanf)
// needs to be checked to assure operation was successful
#include <stdio.h>
#include <stdlib.h>
int main()
{
char choice;
printf("1. 'enter 'A' for this choice\n");
printf("2. 'enter 'B' for this choice\n");
printf("3. 'enter 'C' for this choice\n");
// note
// leading ' ' in format string to consume leading white space
// and no trailing '\n'
if( 1 != scanf(" %c", &choice) )
{ // then, scanf failed
// handle error condition
perror("scanf failed for choice");
exit(EXIT_FAILURE);
}
// implied else, scanf successful
switch (choice)
{
case 'A':
{ // braces needed due to variable declaration
int a = 0; // corrected ':' to ';'
printf("you've chosen menu A\n");
printf("enter a number\n");
// note
// leading ' ' in format string to consume leading white space
// and no trailing '\n'
if( 1 != scanf(" %d", &a) )
{ // then scanf failed
// handle error condition
perror("scanf failed for number");
exit(EXIT_FAILURE);
}
// implied else, scanf successful
printf("%d\n", a);
}
break;
case 'B':
printf("you've chosen B\n");
break;
case 'C':
printf("you've chosen C\n");
break;
default:
printf("your choice is invalid\n!");
break;
} // end switch
return 0;
} // end function: main