Why does scanf ask for input twice, but just in the first loop iteration only?

Question

First of all, this may seem duplicate with this question, but in my case, why did this only happen in the 1st loop iteration (input for the 1st array element). Why not all?

My code:

#include "stdio.h"

int main(int argc, char const *argv[])
{
    int a[5];
    int i;
    for (i = 0; i < 5; i++) {
        printf("Input a[%d]:\n", i);
        int x = scanf("%d ", &a[i]);    // notice the white-space after %d
    }
    for (i = 0; i < 5; ++i)
    {
        printf("a[%d]=%d\n", i, a[i]);
    }
}

Output example:

Input a[0]:
1
2
Input a[1]:
3
Input a[2]:
4
Input a[3]:
5
Input a[4]:
6
a[0]=1
a[1]=2
a[2]=3
a[3]=4
a[4]=5

Why did it ask for input twice only for a[0] but not for the rest, and also why the value assigned to a[1-5] is the value that was input in the one loop iteration before it?

I read this answer, but I still don't understand why it didn't ask for input twice in each loop. Any clear explanation?

`scanf("%d "` --> `scanf("%d"` (drop space). Extra space tells `scanf()` to keep looking until non-white space detected. — chux - Reinstate Monica, Jun 18 '15 at 16:41
please **do not use `scanf()`.** It's extremely hard to use correctly, and it **will** cause subtle, hard-to-track-down bugs. If you want to get one item per line from stdin, use `fgets()` instead. — The Paramagnetic Croissant, Jun 18 '15 at 16:47
@TheParamagneticCroissant: you maybe right about `scanf()` vs `fgets()`, but I'm trying to understand the subtle, hard-to-track-down bugs here. :) — theman, Jun 18 '15 at 16:50
One bug in the code is the returned value from the call to scanf() needs to be properly checked suggest: 'if( 1 != scanf( "%d", &a[i]) ) (// handle error } — user3629249, Jun 18 '15 at 17:46
the compiler raises 3 warnings: 1) unused variable x 2) unused parameter argc 3) unused parameter argv Suggest compiling with all warnings enabled (for gcc, at a minimum, use '-Wall -Wextra -pedantic' ) — user3629249, Jun 18 '15 at 17:48
the code contains the magic number '5' suggest #define MAX_A_LEN (5) then using that name in all instances where '5' currently is written — user3629249, Jun 18 '15 at 17:53

Sourav Ghosh · Accepted Answer · 2015-06-18T16:50:06.817

In your case,

 scanf("%d ", &a[i]);

after scanning an integer, scanf() needs to match the white space, and any number of white space until a non-white space character is required to complete the match.

So, for the first time, the second input is the non-white space, which terminates the scan, but left in the buffer (remains unread) for the next scan.

Next time onwards, the last input (left in the buffer) is considered as the scanned input , and the current input works as the terminator, just to be left in the buffer and so on.

So, the very last input (6), is never actually read into the array, that remains as a mere terminator. The sequential first five inputs are considered. Just to make it clear, quoting from C11, chapter §7.21.6.2, paragraph (5), emphasis mine

A directive composed of white-space character(s) is executed by reading input up to the first non-white-space character (which remains unread), or until no more characters can be read.

Thanks a lot! If only I had known that "the second input is the non-white space, which terminates the scan, but left in the buffer (remains unread) ", I would have been able to answered this my self. So, could you give me some references in standard document or any other article that explain how the input/output buffer works? — theman, Jun 18 '15 at 16:57

score 0 · Answer 2 · edited May 23 '17 at 12:30

@Sourav Ghosh very nicely explained about scanf().

To clarify the user experience, the tricky additional part is that stdin is typically line buffered. That means none of the user typing is given to stdin until Enter or EOF occurs. That is why the first scanf() did not return when 2 was typed (the non-white-space sought by directive " " in scanf("%d ",...). User has to type 2 and Enter, then stdin gets both "2\n". Then scanf() sees the 2, leaves it unread in stdin and then returns.

score 0 · Answer 3 · answered Jun 18 '15 at 17:59

the following code:

1) raises no compiler warnings I.E. complies cleanly

2) properly handles any error condition with scanf()

3) properly handles the program exit with 'return(0);'

4) includes stdlib.h for exit() and EXIT_FAILURE

5) properly declares the main() function when the optional parameters argc and argv are not used.

6) corrects the problem with the format string in scanf()

#include "stdio.h"
#include <stdlib.h> // exit and EXIT_FAILURE

#define MAX_A_LEN (5)

int main( void )
{
    int a[ MAX_A_LEN ];
    int i; // loop index

    for (i = 0; i < MAX_A_LEN; i++)
    {
        printf("Input a[%d]:\n", i);
        if( 1 != scanf("%d", &a[i]) )    
        { // then scanf failed
            perror( "scanf failed" );
            exit( EXIT_FAILURE );
        }

        // implied else, scanf successful
    }

    for (i = 0; i < MAX_A_LEN; ++i)
    {
        printf("a[%d]=%d\n", i, a[i]);
    }

    return(0);
} // end function: main

Why does scanf ask for input twice, but just in the first loop iteration only?

3 Answers3

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