whitespace in the format string (scanf)

Question

Consider the following code:

#include<stdio.h>
int main() {
    int i=3, j=4;
    scanf("%d c %d",&i,&j);
    printf("%d %d",i,j);
    return 0;
}

It works if I give 2c3 or 2 c 3 or 2c 3 as input if I have to change the value of variables. What should I do if I want the user to enter the same pattern as I want means if %dc%d then only 2c3 is acceptable and not 2 c 3 and vice versa if it is %d c %d?

Answer 1

Whitespace in the format string matches 0 or more whitespace characters in the input.

So "%d c %d" expects number, then any amount of whitespace characters, then character c, then any amount of whitespace characters and another number at the end.

"%dc%d" expects number, c, number.

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Also note, that if you use * in the format string, it suppresses assignment:
%*c = read 1 character, but don't assign it to any variable

So you can use "%d%*c c%*c %d" if you want to force user to enter: number, at least 1 character followed by any amount of whitespace characters, c, at least 1 character followed by any amount of whitespace characters again and number.

Answer 2

If you want to accept 1c2 but not 1 c 2, use the pattern without the space:

scanf("%dc%d", &x, &y);

If you want to accept 1c2 and 1 c 2 (and also 1 \t \t c \t 2 etc), use the pattern with the space:

scanf("%d c %d", &x, &y);

If you want to accept 1 c 2 but not 1c2, add a fake string containing whitespace:

scanf("%d%*[ \t]c%*[ \t]%d", &x, &y);

Here the format string %[ \t] would mean "read a string that contains any number of space and tab characters"; but using the additional *, it becomes "expect a string that contains any number of space and tab characters; then discard it"