Question

急求c语言数据结构简单计算器实验报告！！！

简单计算器。
请按照四则运算加、减、乘、除、幂（^）和括号的优先关系和惯例，编写计算器程序。要求：
①从键盘输入一个完整的表达式，以回车作为表达式输入结束的标志。
②输入表达式中的数值均为大于等于零的整数。中间的计算过程如果出现小数也只取整。
例如，输入：4+2*5= 输出：14
输入：(4+2)*(2-10)= 输出：-48
要完整的实验报告。。。

Answer 1

似乎不全面啊!
当年上学的时候编过,优先级比较没这么简单!

Answer 2

#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);
}
}

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);
operand=pop(operand,&operand2);
operator=pop(operator,&op);
operand=push(operand,two_result(op,operand1,operand2));
}
operator=push(operator,expression[position]);
}
else
operand=push(operand,expression[position]-48);
position++;
}
while(!empty(operator))
{
operator=pop(operator,&op);
operand=pop(operand,&operand1);
operand=pop(operand,&operand2);

operand=push(operand,two_result(op,operand1,operand2));
}
operand=pop(operand,&evaluate);
printf("The expression [%s] result is '%d' ",expression,evaluate);
getch();
}

你怎么不去死？