Question

C++模拟计算器的功能，要求至少能实现基本四则运算（加减乘除功能），要求提示明确。

小数也要包括，比如9/2=4.5 尽量精简，希望能把所写代码后面用 /* 写出意思 谢谢大大

Answer 1

#include<iostream>
#include<cmath>
using namespace std;
double main()
{
double m,n;
char ch;
cout<<"请输入两个操作数和运算符:"<<endl;
cin>>m>>n>>ch;
if(ch=='+') cout<<m<<ch<<n<<"="<<m+n<<endl;
else if(ch=='-') cout<<m<<ch<<n<<"="<<m-n;
else if(ch=='*') cout<<m<<ch<<n<<"="<<m*n;
else if(ch=='/') cout<<m<<ch<<n<<"="<<m/n;
return 0;
}

Answer 2

#include <stdlib.h>
#include <ctype.h>
#define MAXOP 100
#define NUMBER '0'
int getop (char [] );
void push (double);
double pop(void);

int getch(void);
void ungetch(int);

int getop(char s[])
{
int i,c;
while ((s[0]=c=getch())==' '|| c=='\t');
s[1]='\0';
if(!isdigit(c)&&c!='.')
return c;
i=0;
if(isdigit(c))
while (isdigit(s[++i]=c=getch()));
if (c=='.')
while (isdigit(s[++i]=c=getch()));
s[i]='\0';
if (c!=EOF)
ungetch(c);
return NUMBER;
}

#define BUFSIZE 100
char buf[BUFSIZE];
int bufp=0;
int getch(void)
{
return(bufp>0) ? buf[--bufp]:getchar();
}
void ungetch(int c)
{
if (bufp>=BUFSIZE)
printf("ungetch:too many characters\n");
else
buf[bufp++]=c;
}
main()
{
int type;
double op2;
char s[MAXOP];
while ((type=getop(s))!=EOF) {
switch (type) {
case NUMBER:
push(atof(s));
break;

case '+':
push(pop()+pop());
break;
case '*':
push(pop()*pop());
break;
case '-':
op2=pop();
push(pop()-op2);
break;
case '/':
op2=pop();
if (op2!=0.0)
push(pop()/op2);
else
printf("error:zero divisor\n");
break;
/*case '/n':
printf("\t%.8g\n",pop());
break;*/
case 0x0a:
printf("\t%.8g\n",pop());
break;

default:
printf("error:unknown command %s\n",s);
break;
}
}
return 0;
}

#define MAXVAL 100

int sp=0;
double val[MAXVAL];

void push(double f)
{
if (sp<MAXVAL)
val[sp++]=f;
else
printf("error:strack full,can't push %g\n",f);
}
double pop(void)
{
if (sp>0)
return val[--sp];
else {
printf("error:stack empty\n");
return 0.0;
}
}

追问

太麻烦了，不用这么多的，能算小数，整数+-*/就ok，代码尽量简单