Input several numbers from array and each one number check for integer or not

Question

everyone! I hope someone can help me figure out something in C language. This is my first seriously homework in IT, I have no experience and I'm learning in e-studies, so teacher help isn't very available.

I need to develop console application in C language. User need to input 10 integer numbers, if insert number isn't integer, need to output error and again re-enter new number until all 10 integer numbers will be inserted.

Everything works in case if I say that these 10 numbers can't be 0 (I make this to be sure that my if-else statement working), but won't work when I want that every input number will be check if it is integer or not.

How can I do it right.

Please help

so far my code look like this:

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

int main()
{
    int i;
    float f;
    int numbers[10];

    for (i = 0; i < 10; i++)
    {
        scanf ("%d", &numbers[i]);
        if (numbers[i] != 0)
        {
            scanf ("*%d", &numbers[i]);
        }
        else
        {
            printf ("\nError!Entered number is't integer \n");
            printf ("\nPlease insert number again \n");
            scanf("%*d", &numbers[i]);
        }

    }

}

see http://stackoverflow.com/questions/11139092/what-happens-when-you-input-things-like-12ab-to-scanfd-argu — Lesto, Mar 20 '14 at 16:59

score 1 · Answer 1 · answered Mar 20 '14 at 17:43

#include <stdio.h>

int main(void) {
    int i = 0;
    int val;
    char ch;
    int numbers[10];

    while(i < 10) {
        val = scanf("%d", numbers + i);  // read the integer into a[i]
        if(val != 1) {
            while((ch = getchar()) != '\n')  // discard the invalid input
                ;  // the null statement
            printf("Error! Entered number is not an integer.\n");
            printf("Please enter an integer again.\n");
            val = scanf("%d", numbers + i);
            continue;
        }
        ++i;
    }
    // process the numbers array
    return 0;
}

I write this line again

val = scanf("%d", numbers + i);

Now it works how I need. Great - thanks a lot

It appears that the 2nd `val = scanf("%d", numbers + i);` should be deleted. — chux - Reinstate Monica, Mar 20 '14 at 20:34

score 0 · Answer 2 · answered Mar 20 '14 at 17:00

0

There are several techniques you might use:

Read the number as a string and reject if it contains characters not suitable for an integer. The use sscanf() to convert the string to integer.
Read the number as a float and reject if it is out of integer range or it has a non-integer value.
Read the input character by character and build up an integer value. If invalid characters appear, reject the value.

answered Mar 20 '14 at 17:00

wallyk

56,922
16
83
148

Thank you for reply so fast. Can you please write me an example how it looks in code for each one of option? Today I was already trying the sscanf thig, but for there somethnig was wrong – effeja Mar 20 '14 at 17:07

ajay · Answer 3 · 2014-03-20T17:48:54.640

scanf returns the number of input items successfully matched and assigned. You can check this value for 1 for each call of scanf. If the value is 0, then you should discard the input to clear the stdin buffer and read input again.

#include <stdio.h>
#include <ctype.h>

int main(void) {
    int i = 0; 
    int val;
    char ch;
    int numbers[10];

    while(i < 10) {
        // read an integer and the first non-numeric character

        val = scanf("%d%c", numbers + i, &ch);  

        // if the number of items assigned by scanf is not 2 or if 
        // the first non-numeric character is not a whitespace, then
        // discard the input and call read input again.
        // for example input of type 32ws are completely discarded

        if(val != 2 || !isspace(ch)) {
            while((ch = getchar()) != '\n')  // discard the invalid input
                ;  // the null statement
            printf("Error! Entered number is not an integer.\n");
            printf("Please enter an integer again.\n");
            continue;
        }
        ++i;
    }
    // process the numbers array
    return 0;
}

Thank you. This is working, but Is there a way that numbers who program eliminated because isn't integer won't counting in array. For example - user input 1 2 3 0,5 (error) 4 (instead of 0,5) 5 6 7 8 9 10 = total input numbers (integer) are 10, because with your code he would not let enter last number because he counts there number who is not integer. — effeja, Mar 20 '14 at 17:29
@effeja Then your best bet is to read the integer as a string and then check if the string contains any non-numeric character. — ajay, Mar 20 '14 at 17:35
@effeja please the updated answer. Now it does what you want. — ajay, Mar 20 '14 at 17:53

score 0 · Answer 4 · answered Mar 20 '14 at 20:12

Although I am not entirely clear on the details of your question, here is an outline of code similar to what you want:

int main(void)
{
    int i;
    int numbers[10];
    int sum = 0;

    for(i=0; i<10; ++i)
    {
        printf("Enter #%d:\n", i+1);
        scanf("%d", numbers+i);

        if (numbers[i] % 2 == 0) // Then Number is even
        {
            sum += numbers[i];
        }
    }

    printf("The sum of only the even numbers is %d\n", sum);

    getch();
    return 0;
}

chux - Reinstate Monica · Answer 5 · 2015-11-19T19:05:28.183

To read an int, suggest fgets() then sscanf() or strtol()

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

int main(void) {
  int i;
  int numbers[10];

  for (i = 0; i < 10; ) {
    char buffer[50];
    if (fgets(buffer, sizeof buffer, stdin) == NULL) break; 
    int n;  // number of `char` parsed
    if (sscanf(buffer, "%d %n", &numbers[i], &n) != 1 || buffer[n] != '\0') {
      printf("Error! Entered number is not an integer.\n");
      printf("Please enter an integer again.\n");
      continue;
    }
    i++;
  }
  return 0;
}

The strtol() approach. This detects overflow issues:

    if (fgets(buffer, sizeof buffer, stdin) == NULL) break; 
    char *endptr;
    errno = 0;  
    long num = strtol(buffer, &endptr, 10);
    if (errno || num < INT_MIN || num > INT_MAX) Handle_RangeError();
    if (buffer == endptr || *endptr != '\n') Handle_SyntaxError();
    numbers[i] = (int) num;

Recommend making a int GetInt(const char *prompt) function that can be used repeatedly.
User input is evil. Do not trust it until well vetted.

Input several numbers from array and each one number check for integer or not

5 Answers5

Linked