CODEWITHSUNDEEP
c
daniel
daniel

Reputation: 637

Why does the code print as much as it gets characters

This is the code

#include <stdio.h>

int main()
{
    int num1 = 0, num2 = 0, num3 = 0, num4 = 0;
    do
    {
        printf("choose 4 numbers, you dont need space\n");
        scanf("%1d%1d%1d%1d", &num1, &num2, &num3, &num4);
        if(!num1 || !num2 || !num3 || !num4)
        {
            getchar();
        }
    }
    while (num1 != num2 || num1 != num3 || num1 != num4 ||
           num2 != num3 || num2 != num4 || num3 != num4); 

}

I don't know what to do with that.

If the input is, for example. */*- then the output is:

choose 4 numbers, you dont need space 
choose 4 numbers, you dont need space 
choose 4 numbers, you dont need space 
choose 4 numbers, you dont need space

Upvotes: 0

Views: 54

Answers (1)

Yennefer
Yennefer

Reputation: 6224

Your code behaves as you describe because the scanf function does not touch the reading buffer when the convertion cannot take place. Calling the function in loop will try to read the next characters as %d.

Entering a non-numeric sequence will never stop your code as expected. For this reason, you should check the return value of scanf in order to detect bad inputs.

Upvotes: 1

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