如何用c语言实现一个计算器？？？

1、#include<stdio.h>int main()

2、{    int a,b,c    scanf("%d%d%d",&a,&b,&c)

3、 int sum = a+b+c

4、   printf("和： %d",sum)

5、printf("平均值：%f",sum/3.0)

6、return 0

讲解：

1、先定义四个整形。

2、一个浮点型保存平均值。

3、然后在控制台等待输入。

4、将输入的三个整数加起来赋值给sum。

5、将三个整形除以3.0（为什么是3.0，是因为ave是浮点型的，隐性转换到float）。

6、然后输出。

#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()

}

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