例3-11 四则运算

例3-11 四则运算

在例3-9、例3-5的基础上增加对除数是否为零的判断
程序核心——if语句

程序

#include<stdio.h>
int main()
{
	double value1,value2;
	char op;
	
	printf("Type in an expression:");
	scanf("%lf%c%lf",&value1,&op,&value2);
	
	if(op=='+')
		printf("=%.2f
",value1+value2);
	else if (op=='-')
		printf("=%.2f
",value1-value2);
	else if (op=='*')
		printf("=%.2f
",value1*value2);
	else if (op=='/')
		if(value2!=0)
			printf("=%.2f
",value1/value2);
		else
			printf("Divisor can not be 0！
"); 
	else
		printf("Unknown operator
");
	return 0;
 }  

结果

Type in an expression:9/0
Divisor can not be 0！

--------------------------------
Process exited after 7.105 seconds with return value 0
请按任意键继续. . .

分析

重点：