例3-11
2013-10-05 09:46 Summer.xia 阅读(107) 评论(0) 编辑 收藏 举报#include<stdio.h> int main(void) { double value1,value2; char op; printf("Type in an expression:"); scanf("%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"); else printf("Unknown operator!\n"); return 0; }