C++ 错误：不允许类型名称
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/15016787/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
C++ Error: Type Name is Not Allowed
提问by user2098000
I'm trying to play around with my new class lesson in Pointer Arguments, and i want to make the functions senior and everyoneElse take pointer x, yet when I try to call the function with the pointer pAge, it says Error: Type name is not allowed. What's wrong?
我正在尝试使用指针参数中的新课程，我想让函数更高级，而everyoneElse 使用指针 x，但是当我尝试使用指针 pAge 调用该函数时，它说错误：类型名称是不允许。怎么了？
#include <iostream>
int senior(int* x);
int everyoneElse(int* x);
using namespace std;
int main()
{
int age(0);
int* pAge(&age);
cout<<"How old are you?"<<endl;
cin>>age;
if(age>59)
senior(int* pAge);
else
everyoneElse(int* pAge);
return 0;
}
int senior(int* x)
{
return *x;
}
int everyoneElse(int* x)
{
return *x;
}
回答by Josh Petitt
if(age>59)
senior(int* pAge);
else
everyoneElse(int* pAge);
You can't include the typename when calling a function. Change to:
调用函数时不能包含类型名。改成：
if(age>59)
senior(pAge);
else
everyoneElse(pAge);
回答by Boyko Perfanov
senior(int* pAge);
else
everyoneElse(int* pAge);
replace with
用。。。来代替
senior(pAge);
else
everyoneElse(pAge);
回答by Anton Kizema
When you call the function, you do not have to specify type of parametr, that you pass to a function:
调用函数时，不必指定传递给函数的参数类型：
if(age>59)
senior(pAge);
else
everyoneElse(pAge);
Parametrs should be specified by type only in function prototype and body function (smth like this:)
参数应仅在函数原型和主体函数中按类型指定（像这样：）
int senior(int* x)
{
return *x;
}
回答by Astro - Amit
How you are calling the function int senior(intx)* and int everyoneElse(intx)* is wrong call the function as : everyoneElse(pAge)and int senior(x)
你如何调用函数int Senior(intx)* 和inteveryoneElse (intx)* 是错误的调用函数为：everyoneElse(pAge)和int Senior(x)
see link http://msdn.microsoft.com/en-us/library/be6ftfba(v=vs.80).aspx
见链接 http://msdn.microsoft.com/en-us/library/be6ftfba(v=vs.80).aspx
