以下代码段中 push(po[i]-'0') 对评估后缀表达式的重要性是什么

Question

实现 C++ 程序，将表达式转换为中缀到后缀，并根据给定条件使用堆栈对其求值操作数和运算符，都必须是单个字符。输入后缀表达式必须采用所需的格式。只需要“+”、“-”、“*”和“/”运算符
我不明白函数 evaluate()

中 push(po[i]-'0') 语句的重要性是什么

#include<iostream>
using namespace std;

class stack
{
    char st[20],in[20],po[20];
    int TOP,k;
public:
    stack()
    {
        TOP=-1;
        k=0;
    }
    void infixToPostfix();
    void evaluate();
private:
    void push(char);
    char pop();
    int precedence(char);
};  

void stack::push(char ch)
{
    if(TOP==19) 
    {
        cout<<"Stack overflow"<<endl;
    }
    else
    {
        TOP++;
        st[TOP]=ch;
    }
}

char stack::pop()
{
    if(TOP==-1)
    {
        cout<<"Stack underflow"<<endl;
        return 0;
    }
    else
    {
        int m=st[TOP];
        TOP--;
        return m;
    }
}

void stack::evaluate()
{
    cout<<"The postfix expression is"<<endl<<po<<endl;;
    int a,b,res,temp;
    TOP=-1;
    for(int i=0;po[i]!='[=10=]';i++)
    {
        if(isdigit(po[i])==1)
        {
            push(po[i]-'0');
        }
        else 
        {
            a=pop();
            b=pop();
            switch(po[i])
            {
                case '+': res=b+a;
                        break;
                case '-': res=b-a;
                        break;
                case '*': res=b*a;
                        break;
                case '/': res=b/a;
                        break;
            }
            push(res);  
        }
    }
    temp=pop();
    cout<<"The answer is "<<temp<<endl;
}

void stack::infixToPostfix()
{
    int m;
    char left='(',right=')';
    cout<<"Enter infix expression"<<endl;       
    cin>>in;
    for(int i=0;in[i]!='[=10=]';i++)        //if operand add it to postfix
    {
        if(isalpha(in[i])==1 || isdigit(in[i]==1))
        {
            po[k]=in[i];
            k++;
        }
        else if(in[i]==left)        //if left parenthesis then push it to stack;
        {
            push(left);
        }
        else if(in[i]==right)       //if right parenthesis encountered then pop from stack until left parenthesis
        {
            while((m=pop())!=left)
            {
                po[k]=m;
                k++;    
            }
        }
        else            //if operator is encounterd pop from the stack the operands having equal or higher precedence
        {
            while(precedence(st[TOP])>=precedence(in[i]))
            {
                int m=pop();
                po[k]=m;
                k++;
            }
            push(in[i]);    
        }
    }
    while(TOP>=0)
    {
        po[k]=pop();
        k++;
    }
    po[k]='[=10=]';
    cout<<"The postfix expression is"<<endl;
    cout<<po;
}

int stack::precedence(char ch)
{
    if(ch=='+' || ch=='-')
    {
        return 1;
    }
    else if(ch=='*' || ch=='/')
    {
        return 2;
    }
    else if(ch=='(')
    {
        return 0;
    }
}

int main()
{
    stack s;
    int op;
    do
    {
        cout<<"\n____________________________"<<endl;
        cout<<"1 Postfix to infix conversion"<<endl;    
        cout<<"2 Evaluation of postfix"<<endl;
        cout<<"3 Exit"<<endl;
        cout<<"______________________________"<<endl;
        cin>>op;
        switch(op)
        {
            case 1: s.infixToPostfix();
                break;
            case 2: s.evaluate();
                break;
            case 3:break;
            default: cout<<"Enter correct option"<<endl;
        }
    }while(op!=3);
    return 0;
}

Answer 1

它将数字字符转换为数字（整数）。例如

if po[i] = '5'.

It pushes, '5'-'0' => 53-48 (their ascii values) = 5.

干杯。

以下代码段中 push(po[i]-'0') 对评估后缀表达式的重要性是什么

what is the importance of push(po[i]-'0') in the following snippet for evaluating postfix expression

c++

postfix