如何用C语言编写一个科学计算器

2024-11-22 04:57:58
推荐回答(5个)
回答1:

用栈 就可以办到了。。。这个很详细的, lz 随便输入一个表达式,中间的计算过程全部输出了,lz试两个 就知道怎么回事了。 #include
#include
#include
#define MAXSIZE 4000;
typedef struct
{
char data[10];
int top;//头地址

int base;//基地址

int length;//长度

}Stack;

void init(Stack *st)//初始化栈

{
st->base=0;
st->top=0;
st->length=0;
}

int isEmpty(Stack *st)
{
int n=0,top,base;
top =st->top;
base =st->base;
if(top==base)
{
return 1;
}
return n;
}

int isFull(Stack *st)
{
int n=0,top,base;
top =st->top;
if(top>=4000)
{
return 1;
}
return n;
}

char getTop(Stack *st)// 返回top值,不改变栈的结构

{
char n;
if(isEmpty(st))
{
printf("栈为空\n");
return 0;
}
int positon= st->top-1;
n= st->data[positon];//取出数据;

return n;
}

char pop(Stack *st)// 出栈,返回

{
char n;
if(isEmpty(st))
{
printf("栈为空\n");
return 0;
}
int positon= st->top-1;
n= st->data[positon];//取出数据;

st->top--;
st->length--;
st->data[positon]='\0';//消除数据

return n;
}

void push(char n,Stack *st)//入栈

{
int positon ;
if(isFull(st))
{
printf("栈满\n");

}
else
{
positon= st->top;//获取位置

st->data[positon]=n;//存入数据

st->top++;//改变位置

}

}

void show(Stack *m1)//输出栈中的数据

{
int top,base;
top=m1->top;
base=m1->base;
while(top>base)
{
printf("%c,",m1->data[--top]);
}
printf("\n");
}

int isOperate(char temp)//是否是操作符

{
if(temp=='+'||temp=='-'||temp=='*'||temp=='/'||temp=='('||temp==')'||temp=='#')
{
return 1;
}
return 0;
}

int isValue(char temp)//是否是数值

{
if(temp>='0'&&temp<='9')//

{
return 1;
}
else
{
return 0;
}
}

int isAvail(char temp)//是否有效字符

{
if(isOperate(temp)||isValue(temp))//如果temp既不是操作符和数值的话,则它是非法的

{
return 1;
}
return 0;
}

int detect(char temp)//搜索矩阵位置

{
int i=0;
char oper[7]={'+','-','*','/','(',')','#'};
for(i=0;i<7;i++)
{
if(temp==oper[i])
{
return i;
}
}
}

char Priority(char temp,char optr)//判断优先级

{
/**//*
+ - * / ( ) #
1 2 3 4 5 6 7
+ 1 < < < < > > >
- 2 < < < < > > >
* 3 > > < < > > >
/ 4 > > < < > > >
( 5 > > > > > = 0
) 6 < < < < = 0 >
# 7 < < < < > 0 =
*/
int row ,col;
char priority[7][7]={/**//* + - * / ( ) # */
{'<','<','<','<','>','>','>'},

{'<','<','<','<','>','>','>'},

{'>','>','<','<','>','>','>'},

{'>','>','<','<','>','>','>'},

{'>','>','>','>','>','=','>'},

{'<','<','<','<','=','0','>'},

{'<','<','<','<','>','<','='},
};

row = detect(temp);//找出对应的矩阵下标;

col = detect(optr);
// printf("%d,%d",row,col);

//优先级存储在一个7x7的矩阵中,对应关系上图;

return priority[row][col];

}
char evaluate(int a,int b,char oper)
{
switch(oper)
{
case '+': return a+b+'0';
case '-': return a-b+'0';
case '*': return a*b+'0';
case '/': return a/b+'0';
default : return 0+'0';
}
}
int calculateExpress(char *express)//计算表达式

{
int result=0;
int a,b;
// char oper,result;

Stack OPTR,OPND;//OPTR存储操作符,OPND操作数值

init(&OPTR);
init(&OPND);
push('#',&OPTR);//默认第一个位'#'

////////////////////-算法-////////////////////////////

while(*express!='\0')
{
char temp;
temp= *(express);
printf("---------------------------------\n");
printf("当前的符号为%c\n",temp);
if(isAvail(temp))//是否是有效字符

{
if(isOperate(temp) )//输入的是操作符

{
char oper,result;
char optr = getTop(&OPTR);//栈中top位的操作符

printf("栈顶操作符位:%c\n",optr);
char prior = Priority(temp,optr);//判断优先级

switch(prior)
{
case '>':
push(temp,&OPTR);
printf("将符号位%c入栈\n",temp);
express++;
break;
case '<':
//int a,b;

//char oper,result;

a=pop(&OPND)-'0';//存在栈中的都是char字符

b=pop(&OPND)-'0';
oper=pop(&OPTR);

result=evaluate(b,a,oper);//出栈一个操作符,计算结果

//printf("%d",result-'0');

push(result,&OPND);//结果入OPND

printf("%d%c%d结果为:%d\n",b,oper,a,result-'0');
break;

case '=':
//消除括号

pop(&OPTR);
printf("消除括号\n");
express++;
break;

}

}
if(isValue(temp))//输入的是数值

{
push(temp,&OPND);//将数值位入栈;

express++;
printf("将数值%c压入栈\n",temp);
//show(&OPND);

}
}
else //表达式中有非法字符

{
printf("表达式中有非法字符\n");
exit(-1);//退出程序

}

}
// show(&OPND);

// show(&OPTR);

return getTop(&OPND)-'0';
}

void inputExpress(char *express)//输入表达式

{
int length=0;
printf("请输入一个表达式:");
scanf("%s",express);
int len =strlen(express);
express[len]='#';//表达式最后一位默认为'#';

express[len+1]='\0';

}

void output(char *express,int result)//输出表达式

{
int i=0;
printf("----------------------------------------\n表达式:");
while(express[i]!='#')
{
printf("%c",express[i]);
i++;
}
printf("=%d\n",result);

}

int main()
{

char express[100];//表达式

int result =0;

inputExpress(express);//输入表达式

result = calculateExpress(express);//计算表达式;

output(express,result); //输出表达式

//、、、、、、、、、、、、、测试优先级。

/**//*
char m='7' ;
m=Priority('+','*');
printf("优先级为%c",m);

int m=evaluate(5,6,'m');
printf("%d",m);
*/
return 0;
}

回答2:

用C语言编写一个科学计算器程序:
#include
void main ()
{float a,b,e;char f;
scanf ("%c",&f);
scanf ("%f,%f",&a,&b);
switch (f){case '+': e=a+b;
printf("%f",d); break;case '-': e=a-b;
printf("%f",d); break;case '*': e = a*b;
printf("%f",d); break;case '/': e=a/b;
printf("%f",d); break;default: printf("Error!"); break;}}

科学型计算器是电子计算器的一种,可进行乘方、开方、指数、对数、三角函数、统计等方面的运算,又称函数计算器。
科学型带有所有普通的函数,所有的函数都分布在键盘上以致于你可以不用通过菜单列表来使用它们。

回答3:

递归计算表达式值,每次从最低优先级算符分割。只需要特殊处理三角函数。

回答4:

用正则表达式的话,很简单。这里给一个以VB.NET为源码的示例。

回答5:

用树和栈的搭配写你这种题目,经典的写法。很好,其实也很好理解的,具体思路531351006

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