What is the effect of trailing white space in a scanf() format string?

Question

What is difference between scanf("%d") and scanf("%d ") in this code, where the difference is the trailing blank in the format string?

#include <stdio.h>  int main(void) {     int i, j;      printf("enter a value for j ");     scanf("%d  ",&j);     printf("j is %d\n", j);     printf("enter a value for i ");     scanf("%d", &i);     printf("i is %d\n", i);     return 0; } 

How does the scanf() function actually work when I add spaces after the format specifier like scanf("%d ", &j);?

Answer 1

A whitespace character in a scanf format causes it to explicitly read and ignore as many whitespace characters as it can. So with scanf("%d ", ..., after reading a number, it will continue to read characters, discarding all whitespace until it sees a non-whitespace character on the input. That non-whitespace character will be left as the next character to be read by an input function.

With your code:

printf("enter a value for j ");  scanf("%d  ",&j);  printf("j is %d \n", j); 

it will print the first line and then wait for you to enter a number, and then continue to wait for something after the number. So if you just type 5Enter, it will appear to hang — you need to type in another line with some non-whitespace character on it to continue. If you then type 6Enter, that will become the value for i, so your screen will look something like:

enter a value for j 5 6 j is 5 enter a value for i i is 6 

Also, since most scanf %-conversions also skip leading whitespace (all except for %c, %[ and %n), spaces before %-conversions are irrelevant ("%d" and " %d" will act identically). So for the most part, you should avoid spaces in scanf conversions unless you know you specifically need them for their peculiar effect.