For loop skipping scanf()

Question

I'm attempting to write a small program where a user specifies they wish to enter n many nunmbers, then they are prompted to enter n many floats in the subsequent lines with a prompt. The expected output is

$./a.out
How many number (at least two)? > 4
Enter a number (3 more to enter) > 2
Enter a number (2 more to enter) > 3
Enter a number (1 more to enter) > 5
Enter a number (0 more to enter) > 7

However, my program currently reads in the first input (2, in the case of my example), then skips the user input of the rest of the prompts.

The output for my current code is

$./a.out
How many number (at least two)? > 4
Enter a number (3 more to enter) > 7
Enter a number (2 more to enter) > 
Enter a number (1 more to enter) > 
Enter a number (0 more to enter) >

I searched for a solution here, here, and here, but I couldn't get anything to work. Does anyone have any suggestions on how to get scanf to work in this? I'd prefer to use scanf as we haven't covered the usage of any of the gets functions.

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

main(){
    int i,n;
    int temp_count,count,rv;
    float number[10000];
    printf("How many number (at least two)? > ");
    rv = scanf("%d",&n);
    if(n<2||rv!=1){
        printf("Please enter at least two numbers");
    }
    count=n;
    for(i=0;i<n;i++){
        count--;
        printf("Enter a number (%d more to enter) > ",count);
        fflush(stdin);
        scanf(" %.2f",number[i]);
    }
    return(0);
}    //end main

Kusalananda · Accepted Answer · 2016-07-14T22:02:28.647

1

As user @user3121023 comments (comment now deleted):

scanf("%f",&number[i]); ampersand needed. Do not fflush(stdin);

The last argument to scanf() should be a pointer to the buffer where the input will be stored. In your case, number[i] is a float, not a float *. Adding a & before number[i] will give scanf() the address of the i:th position of the array.

Also, using fflush() on an input stream is a bad idea as it invokes undefined behaviour. It may work, but it may not. See here for reference on fflush().

Note: fflush() on an input stream is well defined by POSIX (not standard C), and only when the stream is seekable. Standard input is generally not seekable.

edited Jul 14 '16 at 22:02

answered Jul 14 '16 at 21:29

Kusalananda

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Ah, the missing & was a dumb mistake I didn't catch. However, even with fflush(...) removed I'm not getting the correct output. – Jul 14 '16 at 21:32
2

@Zermelo's_Choice Also fix the format specifier. Scan `%f`. You may use `%.2f` with `printf()`, but not with `scanf()`. – Kusalananda Jul 14 '16 at 21:37
1

@Zermelo's_Choice "I'm not getting the correct output" but also accepted answer. Can you clarify what solved your problem so readers will know? – underscore_d Jul 14 '16 at 21:48
@underscore_d Yes, I'll add a note in my original question. – Jul 14 '16 at 22:01
1

@Zermelo's_Choice Thanks, but assuming it was something from this answer, my suggestion would be to write it here, since your question should be... well, a question, not an answer! – underscore_d Jul 14 '16 at 22:02
Firstly I didn't know that the format specifier "%.2f" was only applicable to the printf(...) function. So I changed that to "%f" in the scanf function. Second, I made a silly mistake of not adding the & to 'number[i]' since I was trying to store the input in the ith position of the array. Also, the use of fflush(...) was unnecessary. – Jul 14 '16 at 22:04

For loop skipping scanf()

1 Answers1