第三章例3-11

/* 求解简单的四则运算表达式 */
#include<stdio.h>
int main(void)
{
    double value1,value2;
    char op;

    printf("Type in an expression:");
    scanf_s("%lf%c%lf",&value1,&op,&value2);
    if(op=='+')
        printf("=%.2f\n",value1+value2);
    else if(op=='-')
        printf("=%.2f\n",value1-value2);
    else if(op=='*')
        printf("=%.2f\n",value1*value2);
    else if(op=='/')
        if(value2!=0)
            printf("=%.2f\n",value1/value2);
        else
            printf("Divisor can not be 0!\n");
    printf("Unknown operator!\n");

        return 0;
}