CODEWITHSUNDEEP
c
user130431
user130431

Reputation: 51

Basic C programming

I'm trying to write a program which given a certain number of input will output the product of the listed input (only accounting inputs of 0-9 and ignoring other).

For example: 

input:345 would result output: 60, or another example would be, input: 3t4 and output: 12

I have given it many attempts and this is what I'm stuck with :

#include <stdio.h>


main(){

int c,i;

c = getchar();
i = 1;
while (c!= '\n'){
        if (c>=48 && c<=57){
          i=c*i;
          c=getchar();
        }
}
printf("%d",i);
}

How can i do this ?

Upvotes: 0

Views: 151

Answers (3)

user3251270
user3251270

Reputation: 107

int i,c;

i = 1;

while (i){

    c=getchar();

    if(c == '\n')
    {
        break;
    }

    if (c < 48 || c > 57)
    {
        i = -1;
        break;
    }

    i = i * (c-48);

}

if (i == -1)
{
    printf("Error. Non-Number Entered\n\n");
}

else {

    printf("%d\n\n",i);

}

Upvotes: 0

Dr. Debasish Jana
Dr. Debasish Jana

Reputation: 7118

Use i=(c-48)*i; instead of i = c*i. So, the changed program would be:

#include <stdio.h>

main(){

int c,i;

c = getchar();
i = 1;
while (c!= '$'){
//    printf("%c\n", c);
        if (c>=48 && c<=57){
          i=(c-48)*i;
        }
        c=getchar();
}
printf("%d",i);
}

This will ensure that you are using the numeric value of the digit, 0 not as ascii code of 0 but as simple 0. use c=getchar() outside the if block. This should work.

Upvotes: 0

scarecrow
scarecrow

Reputation: 6864

Couple of issues in your code.

  • After each time the program encounters a non-numeric character it doesn't read further from the input. It reads the same character. Hence c=getchar() should be outside the if block
  • The multiplication happens for the char variable c. It should be converted to the actual number and then be multiplied. In your code you are multiplying its ascii value. Hence (c-48)*i instead of c*i

Upvotes: 1

Related Questions