What's the error in this program code? [closed] - c

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I tried using the ASCII code too by replacing %d with %c in `scanf() code. but then it displays - Add=5 (If I put the input 5 & 6 with 65 ASCII code)
#include <stdio.h>
#include <conio.h>
void main()
{
int num1,num2,res;
char ch;
printf("Enter two numbers");
scanf("%d%d",&num1,&num2);
printf("Enter the operator from the following\n\nA for +\nB for -\nC for*\nD for /\n");
scanf("%c",&ch);
if(ch=='A')
{
res=num1+num2;
printf("Add=%d",res);
}
else if(ch=='B')
{
res=num1-num2;
printf("Sub=%d",res);
}
else if(ch=='C')
{
res=num1*num2;
printf("Mul=%d",res);
}
else if(ch=='D')
{
res=num1/num2;
printf("Div=%d",res);
}
else
{
printf("Invalid Operator");
}
getch();
}
The output:
Any help?

Answer 1

You are facing, sadly, the same problem faced by every beginning C programmer ever: scanf sucks. When you called
scanf("%d%d",&num1,&num2);
it correctly read the two numbers that you typed. But after you typed the two numbers, you probably hit RETURN. And the first call to scanf did not read that return character. So it's still sitting there in the input buffer, and when you later call scanf("%c",&ch), it's actually the newline character that gets read, not the A or B that you wanted to read.
Without changing your program at all, you can try typing something like
12 34A
in response to the first prompt. The first scanf won't read it, so now it will be sitting there in the input buffer waiting for your second call to scanf to find it.
That's a workaround, not a fix. To fix it, you could do what #Ediac suggested in his answer.
(A better fix would probably be not to use scanf at all, but unfortunately that involves learning more about I/O than your book or instructor probably wanted to get into right at first.)

Answer 2

I'd rather write a more useful and correct version of this code
#include <stdio.h>
int main() {
int n1, n2, ans;
char op;
scanf("%d %c %d", &n1, &op, &n2);
switch (op) {
case '+':
ans = n1 + n2;
break;
case '-':
ans = n1 - n2;
break;
case '*':
ans = n1 * n2;
break;
case '/':
if (n2) {
ans = n1 / n2;
break;
}
default:
fprintf(stderr, "Invalid Input\n");
return 1;
}
printf("%d", ans);
return 0;
}
Input
1 + 2
4-3
Output
3
1
See http://ideone.com/y2nnWF demo.