Simple calculator

Problem Statement: Execute simple arithmetic operations sum, subtract, multiplication, division on integers.

How its different from others on internet :  There are so many websites which have code for simple calculator program. However most of them, asks for two integers first and then asks for operator which is not the case in real time calculator. In this code, you can directly enter expressions like        76 * 75, which is similar to real time calculator.

Input : operand1 operator operand2
Output : result of the input arithmetic expression.

Code:

#include <stdio.h>
#include <string.h>

float calculate(int op1, char op, int op2, unsigned short * flagptr); // calculator function
int main()
{
 char oparr1[10], oparr2[10];
 char op;
 float result;
 unsigned short flag = 1;   // flag which indicates any error in input
 unsigned short *flagptr = &flag; // pointer to the above flag
 printf("enter expression operand1 operator operand2\n");
 int ch;
 ch = getchar();  // start reading input
 int i= 0;
 while(ch >= '0' && ch <= '9')   //store in array of operand1 if its number
 {
    oparr1[i] = ch;
    ch = getchar();
    i++;
 }
 while(ch == ' ')       // skip all the spaces between first operand and operator, if any
 {
    ch = getchar();
 }
 op = ch;      // store operator in op
 ch = getchar(); 
 while(ch == ' ')
 {
    ch = getchar();
 }
 i= 0;
 while(ch >= '0' && ch <= '9')      //start reading operand2
 {
    oparr2[i] = ch;
    ch = getchar();
    i++;
 }
 int op1, op2;
 op1 = atoi(oparr1);     // convert char array of operand1 to integer
 op2 = atoi(oparr2);
 result = calculate(op1, op, op2, flagptr);
 if (flag == 1)              // no error in input
    printf("result = %f\n", result);
 else
    printf("undefined result\n");
 return 0;
}

float calculate(int op1, char op, int op2, unsigned short *flagptr)
{
    float result;
    switch(op)
    {
        case '+': result = op1 + op2;
                break;
        case '-': result = op1 - op2;
                break;
        case '*': result = op1 * op2;
                break;
        case '/':
                if (op2 != 0)
                    result = (float)op1/op2;
                else
                    {
                       printf("divide by zero error\n");
                       *flagptr = 0;   // indicate error in input
                    }

                break;
        default : printf("Use integers only\nUse +, -, * and / only\n");
                     *flagptr = 0;
                      break;

    }
    return result;
}
 


Sample input: 25 * 25
Output : 625.00000

What can be improved? 
Ans: Well, this program works only for integers. Try for floating point values and use it as your own calculator. Discuss in comments, how to implement it for float values.