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我们平时进行数学运算都是用计算器完成的，那么如何用C语言编写一个计算器呢？下面我给大家分享一下。
工具/材料Dev C++
01首先我们需要在Dev C++软件中创建一个C语言项目，项目类型选择控制台程序，如下图所示
02接下来我们在项目下面新建C语言文件，如下图所示
03然后我们在C文件中写入计算器逻辑代码，主要是让用户输入计算方式，然后程序自动计算，如下图所示
04接下来我们点击运行菜单，选择下拉菜单中的运行选项，如下图所示
05最后在弹出的界面中我们输入要计算的公式，程序就会自动计算，如下图所示
方法一：
#include <stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
struct complex multiply(struct complex x, struct complex y)
struct complex{
int real
int imag
}
int main()
{
struct complex a,b,s
scanf("%d%d%d%d",&a.real,&a.imag,&b.real,&b.imag)
s=multiply(a,b)
printf("(%d+%di)*(%d+%di)=%d+%di\n",a.real,a.imag,b.real,b.imag,s.real,s.imag)
return 0
}
struct complex multiply(struct complex x, struct complex y)
{
struct complex m
m.real=x.real*y.real-x.imag*y.imag
m.imag=x.imag*y.real+x.real*y.imag
return m
}
方法二：
#include<stdio.h>
int main()
{
int a,b,c,d,e,f
scanf("%d %d %d %d",&a,&b,&c,&d)
e = a * c - b * d
f = a * d + b * c
printf("(%d+%di)*(%d+%di)=%d+%di\n",a,b,c,d,e,f)
}
#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()
}
