CODEWITHSUNDEEP
carraysinteger
effeja
effeja

Reputation: 65

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

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]);
        }

    }

}

Upvotes: 3

Views: 125

Answers (5)

chux
chux

Reputation: 153508

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.

Upvotes: 0

abelenky
abelenky

Reputation: 64682

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;
}

Upvotes: 0

ajay
ajay

Reputation: 9680

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;
}

Upvotes: 0

effeja
effeja

Reputation: 65

#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

Upvotes: 1

wallyk
wallyk

Reputation: 57774

There are several techniques you might use:

  1. 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.

  2. Read the number as a float and reject if it is out of integer range or it has a non-integer value.

  3. Read the input character by character and build up an integer value. If invalid characters appear, reject the value.

Upvotes: 0

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