Multiplication using fgets in C

Question

I am trying to do some multiplication. I am asking users to enter 9 digits number and multiply each digit with. Lets say if user enters 123456789 then we will do this:

123456789
246824682

We will multiply user input with this code:

    #include <stdio.h>
    int main(void) {

        char num[10];
        int num1,num2,num3;

        printf("Enter your num: ");
        fgets(num,10,stdin);

        num1 = num[1] * 2;
        num2 = num[2] * 4;
        num3 = num[3] * 6;

        printf("%d %d %d", num1,num2,num3);



        return 0;
}

When I run this this is what I get:

admin@matrix:~/cwork> ./lab2
Enter your num: 123
100 204 60admin@matrix:~/cwork>

Why am I getting 100 204 and 60 I though I would get : 2 8 18 OR 2818!

Answer 1

First, in the C programming language, arrays are indexed starting at 0. So you need to change the indexes from [1] , [2] , [3] to [0] , [1] , [2] .

Second, each digit in the input is an ASCII character . So the character 2 is actually stored in your array as the number 50. That's why num[1] * 2 is 100.

The easiest way to convert the digits to the numbers you expect is to subtract '0' from each digit. Putting it all together, your code should look like this

num1 = (num[0] - '0') * 2;
num2 = (num[1] - '0') * 4;
num3 = (num[2] - '0') * 6;

Answer 2

When you read from stdin you get a char array.

So in your example with "123"

your array looks like this:

num[1] == '2' == 50 // 50 is the ascii code for '2'
num[2] == '3' == 51 // 51 is the ascii code for '3'

So of course you are getting 100, 204 for these numbers.

Also note that arrays start with 0, not with 1!