Question

谁能用C语言编个完整的程序求表达式的值，例如3*（7-2）。很急！！！谢谢了

要用栈的方法来实现求值。谢谢了

Answer 1

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define TRUE 1
#define FALSE 0

#define LEN 10//输入数字不得超过10位
#define MAXSIZE 40//数字和运算符总个数不得超过40个

typedef struct
{

char data[MAXSIZE][LEN];//栈区为二维数组
int top;

}seqstack;

seqstack *initseqstack();
void push(seqstack *s, char *x);
char *pop(seqstack *s);
int optr(char *z);
int priority(char *a, char *b);
void process(char arrOne[][LEN], char arrTwo[][LEN]);
double calc(char *a, char *b, char *ch);
void express(char (*arrTwo)[LEN]);

seqstack *s = NULL;//声明全局变量

seqstack *initseqstack()//初始化栈顶指针
{
seqstack *s = (seqstack *) malloc (sizeof(seqstack));

s->top = -1;

return s;
}

void push(seqstack *s, char *x)//入栈
{
if (MAXSIZE-1 == s->top)
{
printf("overflow\n");
exit(0);
}

strcpy(s->data[++s->top], x);
}

char *pop(seqstack *s)//出栈
{
char *y = (char *) malloc (LEN * sizeof(char));
strcpy(y, s->data[s->top--]);
return y;
}

int optr(char *z)//判断是否为运算符，是返回TRUE
{
int i;
char *optr1[6] = {"+", "-", "*", "/", "(", ")"};

for (i=0; i<6; i++)
{
if (0 == strcmp(optr1[i], z))
{
return TRUE;
}
}
return FALSE;
}

int priority(char *a, char *b)//比较优先级
{
int i, j;
char *array[5] = {"(", "+", "-", "*", "/"};//优先级列表，由小到大

for (i=0; i<5; i++)
{
if (0 == strcmp(array[i], a))
{
break;
}
}
for (j=0; j<5; j++)
{
if (0 == strcmp(array[j], b))
{
break;
}
}

switch (i)
{
case 0: i = 0; break;
case 1:
case 2: i = 1; break;
case 3:
case 4: i = 2; break;
}

switch (j)
{
case 0: j = 0; break;
case 1:
case 2: j = 1; break;
case 3:
case 4: j = 2; break;
}

if (i >= j)
{
return TRUE;
}
else
{
return FALSE;
}
}

void process(char (*arrOne)[LEN], char (*arrTwo)[LEN])//将中缀式处理成后缀式(运算符入栈处理)
{
int i, j = 0;

s = initseqstack();//初始化

for (i=0; 0!=strcmp(arrOne[i], "\0"); i++)
{
if (1 < strlen(arrOne[i]) || (TRUE != optr(arrOne[i])))//arrOne[i][LEN]为数字时
{
strcpy(arrTwo[j++], arrOne[i]);//将数字拷贝到arrTwo
}
else if (0 == strcmp(arrOne[i], "("))
{
push(s, arrOne[i]);//入栈
}
else if (0 == strcmp(arrOne[i], ")"))
{
while ((-1 != s->top) && (0 != strcmp(s->data[s->top], "(")))
{
strcpy(arrTwo[j++], pop(s));//出栈
}
pop(s);//将"("弹出
}
else//此时arrOne的内容只可能是"+", "-", "*", "/"
{
if (-1 == s->top)
{
push(s, arrOne[i]);
}
else
{
while (-1 != s->top)
{
if (TRUE == priority(s->data[s->top], arrOne[i]))//与栈顶元素进行优先级比较
{
strcpy(arrTwo[j++], pop(s));
}
else
{
push(s, arrOne[i]);
break;
}
}
if (-1 == s->top)
{
push(s, arrOne[i]);
}
}
}
}

while (-1 != s->top)
{
strcpy(arrTwo[j++], pop(s));
}
}

double calc(char *a, char *b, char *ch)//运算，返回double类型
{
double num;
double numOne, numTwo;

numOne = atof(a);
numTwo = atof(b);//将字符串转化为double类型(atof是库函数)

switch(*ch)
{
case '+': num = numOne + numTwo; break;
case '-': num = numOne - numTwo; break;
case '*': num = numOne * numTwo; break;
case '/':
if ((0.000001 >= numTwo) && (-0.000001 <= numTwo))
{
printf("error!\n");
}
else
{
num = numOne / numTwo;
}
break;
default:
exit(0);
}

return num;
}

void express(char (*arrTwo)[LEN])//后缀式入栈运算
{
int i;
double num;

char result[LEN];
char numOne[LEN], numTwo[LEN];

s = initseqstack();//初始化

for (i=0; 0!=strcmp(arrTwo[i], "\0"); i++)
{
if (TRUE != optr(arrTwo[i]))//假如是数，则入栈
{
push(s, arrTwo[i]);
}
else
{
strcpy(numTwo, pop(s));
strcpy(numOne, pop(s));//将顶上两数字弹出

num = calc(numOne, numTwo, arrTwo[i]);//运算
gcvt(num, LEN, result);//将double转换成字符串(gcvt是库函数)

push(s, result);//入栈
}
}

printf(" %s\n\n", result);//输出最终结果
}

void Menu(void)
{
printf("注意:\n");
printf("1.本程序有一定误差，误差范围大约是-0.001到+0.001\n");
printf("2.本程序只能处理正确的表达形式.\n");
printf("3.避免计算结果越界.\n");
printf("4.输入方式:如输入表达式(-1+2)*3\n");
printf(" 输入:\n");
printf(" (空格-1空格+空格2空格)空格*空格3空格=回车\n");
}

void main(void)
{
int i = 0;

char arr[MAXSIZE][LEN]; //输入字符串
char arrOne[MAXSIZE][LEN] = { "\0" }; //保存中缀式
char arrTwo[MAXSIZE][LEN] = { "\0" }; //保存后缀式

Menu();

printf("\n\n\n请输入表达式:\n");

while (scanf("%s", arr[i]))
{
if (0 == strcmp(arr[i], "="))
{
break;
}
strcpy(arrOne[i], arr[i]);
i++;
}
process(arrOne, arrTwo); //调用
express(arrTwo); //调用
}

Answer 2

#include<stdio.h>
void main()
{
printf("%d\n",3*(7-2));
}

追问

要用栈的方法来实现求值。谢谢了

追答

不好意思，我没看到。

Answer 3

#include"stdio.h"
void main()
{int a=3,b=7,c=2;
int d;
d=3*(7-2);
printf("%d\n",d);
}

追问

要用栈的方法来实现求值。谢谢了

Answer 4

#include<stdio.h>
main()
{
int x;//define a variable;
x=3*(7-2);//Expression
printf("x=%d\n",x);//output
return;
}

追问

要用栈的方法来实现求值。谢谢了

Answer 5

我不需要你给分，你可以看下这个资料就行了
表达式运算，我空间里。
中缀转后缀
http://hi.baidu.com/ccocwwh/blog/item/875af00252c529ff08fa938e.html