帮我 编写一个c++风格的计算器程序

答案：
#include <stdio.h> 
struct s_node 
{ int data; 
struct s_node *next; 
};   typedef struct s_node s_list; 
     typedef s_list *link; 
      link operator=NULL; 
      link operand=NULL; 
link push(link stack,int value) 
{ link newnode; 
newnode=(link) malloc(sizeof(s_list)); 
if(!newnode) 
{ printf("\nMemory allocation failure!!!"); 
return NULL; 
} newnode->data=value; 
newnode->next=stack; 
stack=newnode; 
return stack; } 
link pop(link stack,int *value) 
{ link top; 
if(stack !=NULL) 
{ top=stack; 
stack=stack->next; 
*value=top->data; 
free(top); 
return stack; } 
else 
*value=-1; } 
int empty(link stack) 
{ if(stack==NULL) 
return 1; 
else 
return 0; } 
int is_operator(char operator) 
{ switch (operator) 
{ 
case '+': case '-': case '*': case '/': return 1; 
default:return 0; 
} 
} 
int priority(char operator) 
{ 
switch(operator) 
{ 
case '+': case '-' : return 1; 
case '*': case '/' : return 2; 
default: return 0; } } 
int two_result(int operator,int operand1,int operand2) 
{ switch(operator) 
{ case '+':return(operand2+operand1); 
case '-':return(operand2-operand1); 
case '*':return(operand2*operand1); 
case '/':return(operand2/operand1); } } 
答案补充
void main() 
{ char expression[50]; 
int position=0; 
int op=0; 
int operand1=0; 
int operand2=0; 
int evaluate=0; 
printf("\nPlease input the inorder expression:"); 
gets(expression); 
while(expression[position]!='\0'&&expression[position]!='\n') 
{ if(is_operator(expression[position])) 
{ if(!empty(operator)) 
while(priority(expression[position])<= priority(operator->data)&& 
!empty(operator)) 
{ operand=pop(operand,&operand1); 


答案补充
#include<iostream> #include<math.h> using namespace std; main () { char k; double s,g; k='-'; cout<<"求绝对值请按A;求平方根请按B;加减乘除清分别按+-*/" <<endl; cin>>g; while (1) { cin>>k; if (k=='=') break; if (k=='A') { g=fabs (g); break; } if (k=='B') { g=sqrt (g); break; } cin>>s; if (k=='+') g+=s; if (k=='-') g-=s; if (k=='*') g*=s; if (k=='/') { if (s==0) cout<<"wrong"; else g/=s; } } cout<<g; } 

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