非常基本的 C++ 程序问题 - 二进制表达式的无效操作数
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Very basic C++ program issue - Invalid operands to binary expression
提问by Danny
I just started teaching myself C++ on the Mac, and I have run into some issues.
我刚开始在 Mac 上自学 C++，但遇到了一些问题。
I have written some code that allows the user to enter a number and when they hit enter, the number will be returned to the user.
我编写了一些代码，允许用户输入一个数字，当他们按下回车键时，该数字将返回给用户。
Xcode will absolutely not have it though. Every time I try to run my code, it says that there is an issue with the cin>> thisisanumber;code.
Xcode 绝对不会有它。每次我尝试运行我的代码时，它都会说代码有问题cin>> thisisanumber;。
The error comes up and says
错误出现并说
Invalid operands to binary expression.Error is on line 10.
Invalid operands to binary expression.错误在第 10 行。
What am I doing wrong?
我究竟做错了什么？
#include <iostream>
using namespace std;
int main()
{
int thisisanumber();
cout << "Please enter a number: ";
cin >> thisisanumber;
cin.ignore();
cout << "You entered"<< thisisanumber <<"\n";
cin.get();
}
回答by chris
You've fallen victim to the most vexing parse, which means thisisanumberis being treated as a function. Take out the parentheses and you should be fine:
您已经成为最令人烦恼的 parse 的受害者，这意味着thisisanumber被视为一个函数。去掉括号，你应该没问题：
int thisisanumber;
Also consider making it a bit more readable, such as thisIsANumber. If you ever need to know it, thisIsANumberuses the camel-casenaming convention.
还可以考虑使其更具可读性，例如thisIsANumber. 如果您需要了解它，请thisIsANumber使用驼峰命名约定。
回答by Péter T?r?k
Declare your variable without brackets, like
声明没有括号的变量，例如
int thisisanumber;
With brackets, it is interpreted as a function, and a function can't be passed as a parameter to the >>operator.
带括号，它被解释为一个函数，并且一个函数不能作为参数传递给>>操作符。
回答by Grizzly
Your problem is the so called most vexing parse. Basically everything, which could be parsed as a function declaration will be parsed as such. Therefore the compiler will interpret int thisisanumber();as a declaration of a function thisisanumbertaking zero arguments and returning an int. If you consider this behaviour the problems with cin>>thisisanumber;should be somewhat selfevident.
你的问题是所谓的最烦人的 parse。基本上所有可以解析为函数声明的东西都将被解析。因此，编译器将解释int thisisanumber();为一个函数声明，该函数thisisanumber采用零参数并返回一个int. 如果您考虑这种行为，问题cin>>thisisanumber;应该是不言而喻的。
If you remove the parantheses, changing the variable declaration to int thisisanumber;, your program should behave like you'd expect it to with thisisanumberbeing a variable of type int.
如果您删除括号，将变量声明更改为int thisisanumber;，您的程序应该像您期望的那样thisisanumber作为类型的变量int。
You might however reconsider your naming conventions, thisisanumberisn't exactly readable. I would suggest going with this_is_a_number, thisIsANumberor ThisIsANumber.
然而，您可能会重新考虑您的命名约定，thisisanumber这并不完全可读。我建议使用this_is_a_number,thisIsANumber或ThisIsANumber。
回答by kiros hailay
int thisIsANumber;
try making it variable declaration because what you wrote has been interpreted as function.
尝试使其成为变量声明，因为您编写的内容已被解释为函数。
