用简单c语言编写计算器

#include"stdio.h"

专注于为中小企业提供成都网站建设、做网站服务,电脑端+手机端+微信端的三站合一,更高效的管理,为中小企业旅顺口免费做网站提供优质的服务。我们立足成都，凝聚了一批互联网行业人才，有力地推动了数千家企业的稳健成长，帮助中小企业通过网站建设实现规模扩充和转变。

/*预处理命令*/

void

main()

/*主函数*/

{

double

a,b;

/*双精度实型变量说明*/

char

c,d;

/*变量说明*/

do

/*循环体*/

{

printf("input

a

(-*/)b\n");

/*输入提示*/

scanf("%lf%c%lf",a,c,b);

/*输入算术表达式*/

if(c=='

')

/*判断

*/

printf("=%0.2f",a

b);

/*输出a

b的值*/

else

if(c=='-')

/*判断-*/

printf("=%0.2f",a-b);

/*输出a-b的值*/

else

if(c=='*')

/*判断**/

printf("=%0.2f",a*b);

/*输出a*b的值*/

else

if(c=='/')

/*判断/*/

printf("=%0.3f",a/b);

/*输出a/b*/

else

/*不满足以上条件*/

printf("error");

/*输出错误*/

printf("\n\ninput\n");

/*输入\n*/

scanf("%c",d);

/*输入符号给d*/

}

/*循环体结束*/

while(d=='\n');

/*循环条件语句*/

}

C语言编写计算器

总算看懂了，一个只能两个数相加减乘除的计算器何必写的那么复杂，竟然还用了六个函数，下面我写一个功能一样的，更精简方便的，只要一个函数。

/*

Note:Your

choice

is

C

IDE

*/

/*一个具有两个数加减乘除功能的计算器*/

#include

"stdio.h"

void

main()

{

int

iFirNum,iSecNum,iResult;

char

ch,ch1;

printf("请输入表达式如

5+6=

然后按回车键:");

scanf("%d%c%d%c",iFirNum,ch,iSecNum,ch1);

switch(ch)

{

case

'+':

iResult=iFirNum+iSecNum;

printf("%d+%d=%d\n",iFirNum,iSecNum,iResult);

break;

case

'-':

iResult=iFirNum-iSecNum;

printf("%d-%d=%d\n",iFirNum,iSecNum,iResult);

break;

case

'*':

iResult=iFirNum*iSecNum;

printf("%d*%d=%d\n",iFirNum,iSecNum,iResult);

break;

case

'/':

iResult=iFirNum/iSecNum;

printf("%d/%d=%d\n",iFirNum,iSecNum,iResult);

break;

default:

printf("输入表达式错误或该计算器不具备

%ch

功能\n",ch);

}

}

编写函数实现简易计算器的功能（C语言）

#include

void

main()

{

float

a,b;

char

d;

do

{

printf("Please

enter

the

two

Numbers,

separated

by

Spaces:\n");

scanf("%f

%f",a,b);

printf("Please

select

operation

way:

(-,*,/,^,s,!)\n");

scanf("%s",d);

switch(d)

{

case'+':

printf("a+b=%f\n",a+b);

break;

case'-':

printf("a-b=%f\n",a-b);

break;

case'*':

printf("a*b=%f\n",a*b);

break;

case'/':

printf("a/b=%f\n",a/b);

break;

default:

printf("input

error\n");

}

printf("Do

you

want

to

continue(Y/N

or

y/n)");

fflush(stdin);

}

while(toupper(getchar())=='Y');

}

可以运行，不知道满不满足你的要求，你自己可以试试

C语言函数做计算器的问题

在jisuanqi()已经输出，在main()又一次输出jisuanqi()的返回值a+b。可以修改如下:

#include

"stdio.h"

int

jisuanqi(int

a,char

c,

int

b)

{

switch(c)

{

case

'+':

printf("%d\n",a+b);

break;

case

'-':

printf("%d\n",a-b);

break;

case

'*':

printf("%d\n",a*b);

break;

case

'/':

printf("%d\n",a/b);

break;

}

return

0;

}

int

main(int

argc,

char*

argv[])

{

int

a,b;

char

c;

scanf("%d

%c

%d",a,c,b);

jisuanqi(a,c,b);

return

0;

}

用c语言编写计算器

#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();

}

分享题目：c语言计算器函数 c语言计算器函数程序
文章起源：http://csdahua.cn/article/dooscoj.html