Question

我有一个代码，它无法将Bonus读作变量。

这是我的代码：

#include <iostream>
#include <string>
using namespace std;

int main ()
{
    char Bonus,Salary,TotalSalary;
    int num1, num2;

    cout <<"Enter the hours of work for week"<<endl;
    cin >> num1;
    cout <<"Enter your hourly rate"<<endl;
    cin >>num2;

    Salary = num1 * num2;

    if (Bonus >= 45)
        cout <<"Your bonus is 500 pesos"<<endl;
    else if (Bonus>40&&Bonus<=45)
        cout <<"Your bonus is 250 pesos"<<endl;
    else if (Bonus>45&&Bonus<=40)
        cout <<"Your bonus is 150 pesos"<<endl;

    TotalSalary=Salary + Bonus;

    cout <<"Your basic salary is" <<Salary <<"with a bonus of" <<Bonus <<"and a           total of"<<TotalSalary<<endl; 

    system("pause");
    return 0;
}

我希望你能提前帮助我解决我的问题。

Answer 1

因为您的Bonus和Salary以及TotalSalary变量是char。因此，您必须将它们声明为int变量。 Bonus未经初始化即可使用0调用https://en.wikipedia.org/wiki/Undefined_behavior，因此您可以将其初始化为Bonus。此外，您必须要求用户输入cout <<"Enter the Bonus"<<endl; cin >> Bonus;

`#include <iostream>
#include <string>
using namespace std;
int main ()
{
int Salary,TotalSalary;
int num1, num2, Bonus = 0;

cout <<"Enter the hours of work for week"<<endl;
cin >> num1;
cout <<"Enter your hourly rate"<<endl;
cin >>num2;

cout <<"Enter the Bonus"<<endl;
cin >> Bonus;


Salary = num1 * num2;

if (Bonus >= 45)
cout <<"Your bonus is 500 pesos"<<endl;
else if (Bonus>40&&Bonus<=45)
cout <<"Your bonus is 250 pesos"<<endl;
else if (Bonus>45&&Bonus<=40)
cout <<"Your bonus is 150 pesos"<<endl;

TotalSalary=Salary + Bonus;

cout <<"Your basic salary is " <<Salary <<" with a bonus of " <<Bonus <<" and a total of "<<TotalSalary<<endl; 

 system("pause");
return 0;
}

它应该有用。

{{1}}

Answer 2

代码的一些问题：

Salary = num1 * num2;

Bonus，Salary和TotalSalary定义为char，而它们也应定义为int。

Bonus未经初始化使用，会调用undefined behavior。

编译器应该已经警告过你。确保警告已开启且处于最高级别。

有关if statements的一些详细信息：

if (Bonus >= 45)
    cout << "Your bonus is 500 pesos" << endl;
else if (Bonus>40 && Bonus<=45)
    // minor detail, but can't be equal to 45 again
    cout <<"Your bonus is 250 pesos" << endl;
else if (Bonus>45 && Bonus<=40)
    // mayor detail: can never be >45 *and* <=40 at the same time
    cout <<"Your bonus is 150 pesos" << endl;

您可以将其更改为：

// You probably want to check Salary instead of Bonus?
// although the values (like 45) might be off
Salary = num1 * num2;

if (Salary >= 45)
    Bonus = 500;
else if (Salary > 40)
    Bonus = 250;
else
    Bonus = 150;

cout << "Your bonus is " << Bonus << " pesos" << endl;

完整示例：

#include <iostream>
#include <string>

using namespace std;

int main ()
{
    int num1 = 0,
        num2 = 0,
        Bonus = 0,
        Salary = 0,
        TotalSalary = 0;

    cout << "Enter the hours of work for week: ";
    cin >> num1;
    cout << "Enter your hourly rate: ";
    cin >> num2;

    Salary = num1 * num2;

    // values (like 45) might be off here
    if      (Salary >= 45)  Bonus = 500;
    else if (Salary > 40)   Bonus = 250;
    else                    Bonus = 150;

    cout << "Your bonus is " << Bonus << " pesos" << endl;

    TotalSalary = Salary + Bonus;

    cout << "Your basic salary is " << Salary 
         << " with a bonus of "     << Bonus
         << " and a total of "      << TotalSalary << endl; 

    system("pause");
    return 0;
}

Answer 3

HY！定义<?php namespace Easytrip2\Form\Type; use Easytrip2\Form\AbstractEasytrip2Type; use Easytrip2\Form\Select\CountrySelectType; use Easytrip2\Form\Select\GeopointSelectType; use Symfony\Component\Form\AbstractType; use Symfony\Component\Form\FormBuilderInterface; use Symfony\Component\OptionsResolver\OptionsResolver; class StateType extends AbstractEasytrip2Type { /** * * {@inheritDoc} * * @see \Symfony\Component\Form\AbstractType::buildForm() */ public function buildForm(FormBuilderInterface $builder, array $options) { $builder->add ( 'name', 'text', array ( 'label' => 'State name' ) ); $builder->add ( 'code', 'text', array ( 'label' => 'State code' ) ); $builder->add ( 'unloc', 'text', array ( 'label' => 'State unloc code' ) ); // TODO : the validation on this form appears to not be done, thus i try to save (as it is considered as valid) a object which is null, thus fail in the setters. $builder->add ( 'country', new CountrySelectType ( $this->app ), array ( 'label' => false, 'cascade_validation' => true ) ); /** * $builder->add ( 'hub', new GeopointSelectType ( $this->app, 'HUB' ), array ( * 'label' => 'Select a hub if necessary' * ) ); */ } public static function getRefNames() { $return = array (); $countries = CountrySelectType::getRefNames (); // $hubs = GeopointSelectType::getRefNames (); $last; foreach ( $countries as $value ) { $return [] = array ( 'in' => 'state_' . $value ['in'], 'out' => 'state_' . $value ['out'] ); } /* * foreach ( $hubs as $value ) { * $return [] = array ( * 'in' => 'state_' . $value ['in'], * 'out' => 'state_' . $value ['out'] * ); * } */ return $return; } /** * * {@inheritDoc} * * @see \Symfony\Component\Form\AbstractType::configureOptions() */ public function configureOptions(OptionsResolver $resolver) { $resolver->setDefaults ( array ( 'data_class' => 'Easytrip2\Domain\State' ) ); } /** * * {@inheritDoc} * * @see \Symfony\Component\Form\FormTypeInterface::getName() */ public function getName() { return 'state'; } }的位置（类型为char）在

中测试之前声明但从未定义过

Bonus

此时，if(Bonus>=45)可能在运行时给出任何可能的值。没有像Bonus = 0这样的默认赋值。

Answer 4

Bonus没有值，而且是char。

您的变量类型应为int，我怀疑您要将Bonus设置为依赖于Salary的值：

int Bonus = 0;
int Salary = 0;
int TotalSalary = 0;

// ...

if (Salary >= 45)
    Bonus = 500;
else if (Salary > 40)
    Bonus = 250;
else
    Bonus = 150;

TotalSalary = Salary + Bonus;
// ...

无法将Bonus读作变量

4 个答案: