How can I typedef boost::function requiring a type argument?

Asked Jul 16 '14 at 12:23

Active Jul 16 '14 at 15:17

Viewed 195 times

Piecing together some examples I found on stackoverflow, I'd like to do the following:

template<typename T>
using FuncPtr = boost::function< int (const T) >;

Perhaps to later use it like so:

template<typename T>
build_fancy_lookup_table( string key, FuncPtr<T> fp);

However, it doesn't appear to work. I get the following error message:

error: expected unqualified-id before ‘using’

I'm wondering what I'm doing wrong here? I'd like to be able to typedef/alias the boost::function expression to aid in coding.

Here's how I am compiling:

prompt> g++ foo.cpp
prompt> g++ --version
g++ (Debian 4.7.2-5) 4.7.2

thanks.

edited Jul 16 '14 at 15:17

asked Jul 16 '14 at 12:23

Bitdiot

2

`using` aliases are a C++11 feature. – chris Jul 16 '14 at 12:24
1

... so you need to enable C++11 support in your compiler, which might mean using an option like `-std=c++11` – Jonathan Wakely Jul 16 '14 at 12:26
Yes, you guys are right. Thanks. Please post answer. So I can check mark it. I was compiling with g++ no arguments. Let me add that to my post.... – Bitdiot Jul 16 '14 at 12:28

0 Answers0