Scanf is being ignored

Question

#include <stdio.h>
#define length 20
main()
{
    float x;
    int y;
    float array1[length], array2[length], array3[length];
    float ray[length];
    int size1 = insert(array1, length);
    printf("enter number: ");
    scanf("%d", &y);
    int size2 = insert(array2, length);
    int size3 = insert(array3, length);
}

int insert(float a[], int size)
{
    int  n = 0;
    while(n<size && scanf("%f\n", &a[n]) == 1)
    {
        printf("you entered: ");
        printf("%2.1f\n", a[n]);
        n++;
    }
    return n;
}

When I run the program, it executes the first insert okay, but the next time function is called, scanf() seems to be ignored completely. I tried putting it right after where insert is done, but that's ignored as well.

@Olaf Not a duplicate, yet that accepted answer does solve this post. — chux - Reinstate Monica, Oct 04 '15 at 22:06
Your definition of `main()` indicates that you are using a very old book or compiler. — Iharob Al Asimi, Oct 04 '15 at 23:09
What means "the first time"? The first time calling `scanf` or the first time `insert()` with multiple successful `scanf`? What do you enter that `insert()` terminates? If you terminate the input completely, don't wonder that there is no input anymore in further calls... A little dump of the console output would be helpful — Daniel Alder, Oct 04 '15 at 23:16
1 you entered: 1.0 enter new number 2 you entered: 2.0 enter new number 3 you entered: 3.0 enter new number 4 you entered: 4.0 enter new number 5 you entered: 5.0 enter new number 6 you entered: 6.0 enter new number k enter number: after inserting a non-number it will stop — sajax63, Oct 04 '15 at 23:30
do not use `scanf`. Get the whole line with `fgets` and the parse it with `sscanf`. You have much more control with these two functions — Pablo, Oct 04 '15 at 23:40
@sajax63 you should show the unexpected output in the question, instead of as a comment — M.M, Oct 05 '15 at 00:06

anurag-jain · Accepted Answer · 2015-10-05T00:14:03.793

Use %*c in scanf to consume the newlines along with space around %d in the scanf in main(). I tested the below code on MingW and it seem to work. The '\n' in your scanf is being consumed making it scanf() return while the '\n' at the press of enter key still remains in IO buffer to be consumed by scanf() again; hence the weird behaviour.

#include <stdio.h>
#include <stdlib.h>

#define length 20

int insert(float *a, int size)
{
    int  n = 0;
    while(n<size && scanf("%f%*c", &a[n]))
    {
        printf("you entered: ");
        printf("%2.1f\n", a[n]);
        n++;
    }
    return n;
}

int main(int argc, char* argv[])
{
float x;
    int y;
    float array1[length], array2[length], array3[length];
    float ray[length];
    int size1 = insert(array1, length);
    printf("enter number: ");
    scanf("%d", &y);
    int size2 = insert(array2, length);
    int size3 = insert(array3, length);
    return 0;    
}

Why? I did quick run on DevC++ and it doesn't complain. – anurag-jain Oct 05 '15 at 00:08 — anurag-jain, Oct 05 '15 at 00:08
@M.M Works fine with that as well. – anurag-jain Oct 05 '15 at 00:11 — anurag-jain, Oct 05 '15 at 00:11
Cool, will remove my comments now that it is fixed – M.M Oct 05 '15 at 00:34 — M.M, Oct 05 '15 at 00:34

score 0 · Answer 2 · answered Oct 05 '15 at 00:05

In the scanf format string, change "%f\n" to "%f". The \n in a scanf format string does not mean "wait for newline".

You do not need to worry about waiting for newline, because the only format specifiers you use are %f and %d, which both discard any leading whitespace.

Scanf is being ignored

2 Answers2

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