CODEWITHSUNDEEP
c
friends
friends

Reputation: 639

How to accept 3 arguments as int

For example:

console> please enter 3 digits: 1 2 3

I only know how to accept 1 digit using scanf:

scanf("%d", &space);

Upvotes: 0

Views: 203

Answers (2)

paxdiablo
paxdiablo

Reputation: 882078

Have you tried:

scanf("%d %d %d", &num1, &num2, &num3);

That will work fine if you know you want just the three. If you want a variable number, you'll need to do it in a loop, something like:

#include <stdio.h>

int main (void) {
    int i, ch, rc, val, count;

    // Loop until we get a valid number.

    rc = 0;
    while (rc != 1) {
        printf ("Enter count: "); fflush (stdout);
        rc = scanf(" %d", &count);

        // Suck up all characters to line end if bad conversion.

        if (rc != 1) while ((ch = getchar()) != '\n');
    }

    // Do once for each number.

    for (i = 1; i <= count; i++) {
        rc = 0;
        while (rc != 1) {
            printf ("\nEnter #% 2d: ", i); fflush (stdout);
            rc = scanf(" %d", &val);
            if (rc != 1) while ((ch = getchar()) != '\n');
        }
        printf ("You entered %d\n", val);
    }

    return 0;
}

Running this gives you:

Enter count: 5

Enter # 1: 10
You entered 10

Enter # 2: 20
You entered 20

Enter # 3: 30
You entered 30

Enter # 4: 40
You entered 40

Enter # 5: 99
You entered 99

Upvotes: 4

Brian Clements
Brian Clements

Reputation: 3905

You can read in multiple numbers with scanf

int a, b, c;
scanf("%d %d %d", &a, &b, &c);

Upvotes: 8

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