C语言结课设计之计算器功能

本文实例为大家分享了C语言实现计算器的具体代码，供大家参考，具体内容如下

 
#include 
#include 
#include 
#define maxn 1000
char buf[maxn], str[maxn], signStack[maxn], ch[2];
int len, id, idSign, idAns, i, n;
double ans[maxn];
 
void checkSign(char sign){
 if(sign == '(') signStack[idSign++] = sign;
 else if(sign == '*' || sign == '/'){
 while(idSign && (signStack[idSign-1] == '*' || signStack[idSign-1] == '/'))
 str[id++] = signStack[--idSign];
 signStack[idSign++] = sign;
 }else if(sign == ')'){
 while(signStack[idSign-1] != '(') str[id++] = signStack[--idSign];
 --idSign;
 }else{
 while(idSign && signStack[idSign-1] != '(') str[id++] = signStack[--idSign];
 signStack[idSign++] = sign;
 }
 str[id++] = ' ';
}
 
int check(double a, char sign){
 if(a == 0 && sign == '/'){
 printf("除数不能为0!，程序结束。n");
 system("pause");
 exit(EXIT_FAILURE);
 }
 return 1;
}
 
double cal(double a, double b, char sign){
 switch(sign){
 case '+': return a + b;
 case '-': return a - b;
 case '*': return a * b;
 case '/': return a / b;
 }
}
 
int main(){ 
 
 printf("tttt简单计算器nn");
 printf("tt ======================================n");
 printf("tt || 支持+ - * / ( ) 等符号和负数输入 ||n");
 printf("tt ======================================n");
 printf("tttttt原创作者：邱正钢n");
 printf("tttttt完成时间：2014年6月17日n");
 printf("ttttttEmail：2276479303@qq.comn");
 
 printf("请输入一个表达式，如 -10*(20.2-30.6)+5.0/2n以回车结束:n"); 
 do{
 gets(buf); id = idSign = idAns = 0;
 if((len = strlen(buf)) == 0) continue;
 
 int flag = 0; //这个是用来判断'-'字符是负号还是减号的。1表示数字，0表示 +*/(
 
 
 for(i = 0; i < len; ++i){
 if(buf[i] == ' ') continue;
 if(buf[i] >= '0' && buf[i] <= '9' || buf[i] == '.' || buf[i] == '-' && !flag){
 str[id++] = buf[i]; flag = 1;
 }
 else { 
 str[id++] = ' '; checkSign(buf[i]); 
 if(buf[i] != ')') flag = 0;
 else flag = 1;
 }
 }
 while(idSign) str[id++] = signStack[--idSign];
 
 for(i = 0, n = 0; i < id; ++i){
 if(str[i] == ' ') continue;
 
 if(str[i] >= '0' && str[i] <= '9' || str[i] == '.' || str[i] == '-' 
 && (str[i+1] >= '0' && str[i+1] <= '9' || str[i+1] == '.')){
 sscanf(str + i, "%lf%n", &ans[idAns++], &n);
 i += n - 1; continue;
 }else if(check(ans[idAns-1], str[i])){
 ans[idAns-2] = cal(ans[idAns-2], ans[idAns-1], str[i]);
 --idAns; 
 }
 }
 printf("结果是 %.2lfn", ans[0]);
 printf("输入Y 继续, N 退出: ");
 scanf("%s", ch);
 }while(ch[0] == 'Y' || ch[0] == 'y');
 printf("感谢您的使用！再见n");
 system("pause");
 return 0;
}

关于计算器的精彩文章请查看《计算器专题》 ，更多精彩等你来发现！

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