问题描述

错误 1 error C2872: “less”: 不明确的符号

#include "stdafx.h"
#include "stdio.h"
#include "string.h"
#include "ctype.h"
#include "malloc.h"
#include
using namespace std;

#define N 1000

typedef struct input *Pin;//表达式链结点
struct input
{
int info;//数字用数值存储，符号用对应的ASCII码存储
int type;//标记：0为数字，1为符号
Pin next;
}head;

typedef struct node *pnode;//栈结点
struct node
{
int info; //结点内容
pnode next;
};

struct LinkStack
{
pnode top;
};
typedef struct LinkStack * PLinkStack;

PLinkStack OPTR, OPND;//符号栈、数值栈
Pin E;
Pin now;
char in[N];

int top(PLinkStack plstack);//取栈顶元素
void push(PLinkStack plstack, int n);//入栈
int pop(PLinkStack plstack);//出栈
int signature(int n);//符号处理模块
int transform(int c);//将运算符转换成符号分析表中的坐标
int cmp(int i, int j);//查符号分析表，返回优先级
int less(Pin E);//符号表小于操作
int equal(Pin E);//符号表等于操作
int greater(Pin E);//符号表大于操作
void change();////将输入字符串转化为链表，在输入表达式前后各加一个#
void start();//获得输入字符，检测其合法性
Pin provide();//每次调用时返回一个结点

int main()
{
int flag1 = 1, flag2 = 1, temp, value;
OPND = (struct LinkStack *) malloc(sizeof(struct LinkStack));
OPTR = (struct LinkStack *) malloc(sizeof(struct LinkStack));

while (flag1)//控制可以循环输入表达式
{
    flag2 = 1;
    OPND->top = NULL;
    OPTR->top = NULL;
    push(OPTR, '#');

    start();//获得输入字符，检测其合法性
    printf("后缀表达式：");

    while (flag2 != 0)
    {
        E = provide();//获得一个结点，数字/符号
        if (E->type == 0)//数字处理
        {
            printf("%d ", E->info);//输出后缀式子控制
            push(OPND, E->info);
        }
        else if (E->type == 1)//符号处理
        {
            temp = signature(E->info);
            if (temp == 1)
            {
                value = pop(OPND);//表达式的值
                flag2 = 0;
            }
        }
        else
        {
            printf("无法获得结点信息！n");
        }
    }
    printf("n表达式的值为：%dn", value);
    printf("n");
}

}

void start()//获得输入字符，检测其合法性
{
head.next = NULL;
now = &head;
int i;
int len;
bool pass = true;
int flag = 0;

while (pass)
{
    printf("请输入表达式：");
    scanf_s("%s", in);

    len = strlen(in);

    for (i = 0; i<len; i++)//输入表达式合法检测：只能输入数字、+、-、*、/、(、)。对于")(" "i("v ")i" 格式报错
    {
        if ((in[i] >= '0'&&in[i] <= '9') || (in[i] == '+') || (in[i] == '-') || (in[i] == '*') || (in[i] == '/') || (in[i] == '(') || (in[i] == ')'))
        {
            if (!(in[i] >= '0'&&in[i] <= '9') && !(in[i + 1] >= '0'&&in[i + 1] <= '9'))
            {
                if (!(in[i] == ')' || in[i + 1] == '('))
                {
                    flag = 1;
                    printf("表达式有误!nn");
                }
                if (in[i] == ')'&&in[i + 1] == '(')
                {
                    flag = 1;
                    printf("表达式有误!nn");
                }
            }
            else if ((in[i] >= '0'&&in[i] <= '9'&&in[i + 1] == '(') || (in[i] == ')'&&in[i + 1] >= '0'&&in[i + 1] <= '9'))
            {
                flag = 1;
                printf("表达式有误!nn");
            }
        }
        else
        {
            flag = 1;
            printf("错误：非法字符!nn");
        }
    }
    if (flag == 1)
    {
        flag = 0;
        pass = true;
    }
    else
        pass = false;
}
change();

}

void change()//将输入字符串转化为链表，在输入表达式前后各加一个#
{
int num = 0;
char *p, *q;
p = &in[0];
q = &in[1];

while (*p != '')
{
    if (*p >= '0'&&*p <= '9')
    {
        num = num * 10 + (*p - '0');
        if (!(*q >= '0'&&*q <= '9'))
        {
            Pin temp = (Pin)malloc(sizeof(struct input));
            temp->next = NULL;
            now->next = temp;
            now = temp;
            temp->info = num;
            temp->type = 0;
            num = 0;
        }
    }
    else
    {
        Pin temp = (Pin)malloc(sizeof(struct input));
        temp->next = NULL;
        now->next = temp;
        now = temp;
        temp->info = *p;
        temp->type = 1;
    }
    p++;
    q++;
}
Pin temp = (Pin)malloc(sizeof(struct input));
temp->next = NULL;
temp->info = '#';
temp->type = 1;
now->next = temp;
now = head.next;

}

Pin provide()//每次调用时返回一个结点
{
Pin temp = now;
now = now->next;
return temp;
}

int top(PLinkStack plstack)//取栈顶元素
{
return plstack->top == NULL ? -1 : plstack->top->info;;
}

void push(PLinkStack plstack, int n)//入栈操作
{
pnode tem = (struct node *)malloc(sizeof(struct node));
tem->info = n;
tem->next = NULL;
if (plstack->top == NULL)
{
plstack->top = tem;
return;
}
tem->next = plstack->top;
plstack->top = tem;
return;
}

int pop(PLinkStack plstack)//出栈操作
{
int tem;
if (plstack->top == NULL)
return -1;
tem = plstack->top->info;
plstack->top = plstack->top->next;
return tem;
}

int signature(int n)
{
int i, j, temp;//i前一个符号，j后一个符号
int comp;
if (top(OPTR) == -1) { push(OPTR, n); }//栈为空，直接压入栈
else
{
comp = cmp(top(OPTR), n);//比较运算符优先级
}
switch (comp)
{
case -1:temp = less(E); break;
case 0:temp = equal(E); break;
case 1:temp = greater(E); break;
}
if (n == 35 && top(OPTR) != 35)//！！！当前符号为表达式末尾的#，若符号栈内有未计算的符号，递归调用符号处理
{
temp = signature(n);
}
return temp;
}

int less(Pin E)//当前符号进栈OPTR
{
push(OPTR, E->info);
return 0;
}

int equal(Pin E)
{
int z;
if (E->info == 35) return 1;//当前符号为表达式末尾的#
else
{
z = pop(OPTR);
return 0;
}
}

int greater(Pin E)//基于当前符号的计算
{
int a, b, t, r, flag = 1, Z;
b = pop(OPND);
a = pop(OPND);
t = pop(OPTR);
printf("%c ", t);
switch (t)
{
case 43:r = a + b; break;//'+'
case 45:r = a - b; break;//'-'
case 42:r = a*b; break;//'*'
case 47:r = a / b; break;//'/'
}
if (E->info == ')') //运算结束后将'('出栈
pop(OPTR);
else
{
if (E->info != '#')
push(OPTR, E->info);
}
push(OPND, r);

return E->info == 35 ? 1 : 0;//是#，分析结束，输出结果；不是#，继续取E

}

int cmp(int i, int j)//查表并返回表中优先级，<为-1，=为0，>为1，-2为出错
{
int first, second;
int Table[7][7] = { { 1, 1, -1, -1, -1, 1, 1 }, { 1, 1, -1, -1, -1, 1, 1 }, { 1, 1, 1, 1, -1, 1, 1 }, { 1, 1, 1, 1, -1, 1, 1 }, { -1, -1, -1, -1, -1, 0, -2 }, { 1, 1, 1, 1, -2, 1, 1 }, { -1, -1, -1, -1, -1, -2, 0 } };
return Table[transform(i)][transform(j)];
}

int transform(int ch)//将符号转换为分析表中的坐标
{
int num;
switch (ch)
{
case 43:
num = 0; break;
case 45:
num = 1; break;
case 42:
num = 2; break;
case 47:
num = 3; break;
case 40:
num = 4; break;
case 41:
num = 5; break;
case 35:
num = 6; break;
default:
printf("符号%c错误！", ch);
}
return num;
}