So, I just want to make this function:
template<typename T>
void printWithEndl(T)
{
std::cout << T << "\n";
}
but I got this error on the line:
std::cout << T << "\n";
I wonder: how can I cout the value of T?
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Peter Mortensen
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JT Woodson
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2You forgot to give the variable a name. The type alone is not enough. – NathanOliver Dec 22 '16 at 17:52
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You could do `typeid(T).name()` but it's not necessarily going to give you anything useful. – Edward Strange Dec 22 '16 at 17:54
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I think he wants to print the type used in that instance of the template. "int", "char", or whatever was used for instantiating. Not sure if possible in standard C++. – Desaroll Dec 22 '16 at 18:06
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Also you'd have to overload the << operator for any type you pass there in. – germanfr Dec 22 '16 at 18:11
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Oh, I cannot believe I made such a rookie mistake, thanks for pointing that out. – JT Woodson Dec 22 '16 at 18:36
1 Answers
10
You should name the variable you're passing to printWithEndl, and cout that name:
template<typename T>
void printWithEndl(T msg)
{
std::cout << msg << "\n";
}
If you're using this to print complex objects, you're probably better off passing a reference to const:
template<typename T>
void printWithEndl(const T& msg)
{
std::cout << msg << "\n";
}
Jack Deeth
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1Differing variable and type names by case is considered a bad programming practice. – Thomas Matthews Dec 22 '16 at 18:13
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1In your answer, you have a data type `T` and a variable `t`. One of these should be a different name. – Thomas Matthews Dec 22 '16 at 20:51
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