错误:转换为非标量类型

art*_*00n 2 c++ cstring

我正在为一个赋值创建一组派生类.我被指示使用char数组(c-strings).当我编译时,我不断收到错误:

Homework11.cpp: In function âint main()â:
Homework11.cpp:72: error: conversion from âchar [10]â to non-scalar type âBusinessâ requested
Homework11.cpp:73: error: conversion from âchar [10]â to non-scalar type âBusinessâ requested
Homework11.cpp:74: error: conversion from âchar [10]â to non-scalar type âAccountâ requested
Homework11.cpp:75: error: conversion from âchar [10]â to non-scalar type âAccountâ requested
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我相当肯定我的问题源于我尝试将实例变量Name设置为发送的参数.这是我的代码,其中包含注释,我相信问题可能存在.

#include <iomanip>
#include <iostream>
#include <cstdlib>
#include <cstring>
using namespace std;

class Person{
public:
        Person() {}
        Person(char theName[]) {strcpy(name,theName);}
        void getName(char theName[]) // I think the problem may be here or in the line above
                { theName = name;}
private:
        char name[80];
 };

class Account : public Person{
public:
        Account() :accountNum(0),balance(0) {}
        Account(int actNo, char theName[])
                :Person(theName),accountNum(actNo),balance(0) {}
        void setBal(float theBalance)
                {balance = theBalance;}
        void deposit(float numDeposited)
                { balance = balance + numDeposited;}
        float withdraw(float numWithdrawn)
                { balance = balance -numWithdrawn;
                  return numWithdrawn;}
        float getBal() {return balance;}
        void printBal();
private:
        int accountNum;
        float balance;
};

 class Business : public Account{
 public:
        Business() : checkFee(0.0) {}
        Business(int actNo, char theName[])
                : Account(actNo, theName),checkFee(0.0) {}
        float withdraw(float numWithdrawn)
                {float newBalance = getBal()-numWithdrawn-checkFee;
                 setBal(newBalance);
                  return numWithdrawn;}
        void setFee(float fee) {checkFee = fee;}
 private:
        float checkFee;
};

void Account::printBal()
{
        char name[80];
        getName(name);
        cout<<setw(10)<<"Account # "<<accountNum<<setw(10)<<
              name<<setw(10)<<balance<<endl;
}


int main()
{
        char businessName1[10]="Business1";
        char businessName2[10] ="Business2";
        char regularName1[10] = "Regular1";
        char regularName2[10] = "Regular2";

       //The following 4 lines are the ones I am getting the error for
        Business bs1 = (1,businessName1);
        Business bs2 = (2,businessName2);
        Account rg1 = (1, regularName1);
        Account rg2 = (2, regularName2);

        cout<<"Intially: "<<endl;
        rg1.printBal();
        rg2.printBal();
        bs1.printBal();
        bs2.printBal();

        bs1.deposit(1000.00);
        bs2.deposit(1000.00);
        rg1.deposit(1000.00);
        rg2.deposit(1000.00);

        cout<<"----------------------------------------"<<endl;
       cout<<"After adding 1000.00 to all accounts:"<<endl;
        rg1.printBal();
        rg2.printBal();
         bs1.printBal();
        bs2.printBal();

        bs1.setFee(1.00);
        bs1.withdraw(500);
        bs2.withdraw(500);
        bs1.deposit(250);
        bs2.deposit(250);
        rg1.withdraw(500);
        rg2.deposit(500);

        cout<<"---------------------------------------"<<endl;
        cout<<"Finially:"<<endl;
        rg1.printBal();
        rg2.printBal();
        bs1.printBal();
        bs2.printBal();

        return 0;
}
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Jes*_*ood 8

正确的语法是Business bs1(1,businessName1);.如果要使用=,还可以使用复制初始化Business bs2 = Business(2,businessName2);.

前者称为直接初始化.它们并不完全相同,请参阅复制初始化和直接初始化之间的C++是否存在差异?有关深入的信息.

Business bs1 = (1,businessName1);所述1和阵列businessName1由分离逗号运算符.逗号运算符计算第一个操作数,即1抛弃结果并返回第二个操作数的值,这是您的情况下的数组.换句话说,您的代码相当于Business bs1 = businessName1;.这就是错误消息说它无法将a转换char[10]Business对象的原因.