Very simple two-operand expression 'parser'

#include <stdio.h>

int calc(int op1, int op2, char op);
short op_v(char op);

int main(void)
{
    /* Variable declarations */
    char op;
    int op1, op2;
    
    /* Get the expression */
    printf("%s", "Enter a simple expression: ");
	
    /* Break the expression down into tokens and solve it */
    if(scanf("%d%c%d", &op1, &op, &op2) == 3 && op_v(op))
	printf("%s%d", "Result= ", calc(op1, op2, op));
    else
	printf("%s", "Invalid expression.\n");
    
    return 0;
}

/* Calculate the result */
int calc(int op1, int op2, char op)
{
    switch(op)
    {
    case '+': return op1+op2;
    case '-': return op1-op2;
    case '*': return op1*op2;
    case '/': return op1/op2;
    }
}

/* Validate the operator */
short op_v(char op)
{
    switch(op)
    {
    case '+': return 1;
    case '-': return 1;
    case '*': return 1;
    case '/': return 1;
    default: return 0;
    }
}