Question

我正在研究这个程序，但是我收到2个错误，我不知道它们是什么意思或我如何修复它们.... 程序应该读取任何等式为+ b-c或（a + b）-c ......等并检查括号是否匹配这是代码

// A simple Integer stack, Array based
#include <string>
#include <iostream>
#define STACK_MAX 100
#define equation

using namespace std;
template <typename type>
class Stack {

private:
    type        data[STACK_MAX];
    int        size;

public:
    Stack();
    ~Stack() { }   // Destructor
    int top();
    void push(int d) ;
    void pop() ;
    int is_full ();
    int is_empty () ;
    void equal( Stack& entry) ;
};
template <typename type > int Stack<type>::is_full () {
    return ( STACK_MAX >= size );
}
template <typename type > int Stack<type>::is_empty () {
    return ( STACK_MAX == 0 );
}
template <typename type>  Stack <type>::Stack() {
    size = 0;
}
template <typename type > int Stack< type >:: top() {

    if (is_empty ()) {
        printf("Error: stack empty\n");
        return -1;
    }
    return data[size-1];
}
template <typename type > void Stack <type>::push (int d) {

    if (! is_full () ) {
        data[size++] = d;
    } else {
        printf("Error: stack full\n");
    }
}
template < typename type > void Stack <type>::pop() {

    if (size == 0) {
        printf("Error: stack empty\n");
    } else {
        size--;
    }
}
template <typename type > void Stack <type> ::equal ( Stack& entry ) {
    std::string& equation ;
    char left = '(' ;
    char right= ')' ;
    Stack <type> x ;
    string::size_type a ;
    char next ;
    next = entry ;
    bool failed = false ;
    for ( a= 0 ; ! equation.length() ; ++a) {
        next = equation[a] ;
        if ( next == left ) {
            x.push (next);
        } else {
            if (x.is_empty()) {
                x.pop () ;
            } else {
                failed = true ;
            }
        }

    }
    if ( a == 0) {
        cout << " No parenthesis"<< endl ;
    } else if ( x.top () == left ) {
        cout << "Parenthesis don’t match. Missing right parenthesis" << endl ;
    } else if ( x.top () == right ) {
        cout << "Parenthesis don’t match. Missing left parenthesis " << endl ;
    } else {
        cout << "Matching parenthesis" << endl ;
    }

}


int main() {

    Stack <char> equation2 ;
    char b ;
    cout << " please enter your equation " << endl ;
    for (int i =0 ; i< 100 ; i++ ) {

        cin >> b ;
        equation2.equal(b) ;
    }


    system ("pause") ;


    return 0;
}

这是错误

    2   IntelliSense: a reference of type "Stack<char> &" (not const-qualified) cannot be initialized with a value of type "char"   c:\users\mike\documents\visual studio 2010\projects\stack\stack\stack.cpp   101
   Error    1   error C2664: 'Stack<type>::equal' : cannot convert parameter 1 from 'char' to 'Stack<type> &'   c:\users\mike\documents\visual studio 2010\projects\stack\stack\stack.cpp   101

Answer 1

之前的答案因某种原因被删除了，所以我想我会发布一个。

正如你所说，equ当前的实现取决于要处理的类型是char，它会将签名更改为type＆amp;没有意义，因为如果使用char类型实例化模板化类，它仍然只能按预期工作。然而，目前的问题是等号签名需要Stack＆amp; amp;这是错误的，你应该改为char

template <typename type > void Stack <type> ::equal ( Stack& entry )

到

template <typename type > void Stack <type> ::equal ( char& entry )

错误C2664：'Stack <type> :: equal'：无法将参数1从'char'转换为'Stack <type>＆amp;'</type> </type>

1 个答案: