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Use of generic function with std::transform

Posted on 2013-12-18
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Last Modified: 2013-12-24
Suppose I write a generic function to calculate the square of a number

template<class T> T square(const T & in)
{
	return in*in;
}

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and I have a vector of integers called v. When I try to run the following code, I get an error:

std::vector<int> v2;
std::transform(v.begin(), v.end(), std::back_inserter(v2), square);

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I get an error. However, if I create the following struct

struct squareTempl
{
	template< typename T >
	T operator ()(const T& in) const {return in*in;}
};

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then I can use

std::transform(v.begin(), v.end(), std::back_inserter(v2), squareTempl());

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without any problems. Why is there such a discrepancy?
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Question by:Rothbard
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3 Comments
 
LVL 86

Assisted Solution

by:jkr
jkr earned 125 total points
ID: 39727499
That's not really a discrepancy, 'transform()' (http://www.sgi.com/tech/stl/transform.html) requires a class/struct modeled after 'UnaryFunction'  (http://www.sgi.com/tech/stl/UnaryFunction.html) for this transformation. And a 'UnaryFunction' function object (http://www.sgi.com/tech/stl/functors.html) needs to provide an overloaded 'operator ()'.
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LVL 40

Accepted Solution

by:
evilrix earned 125 total points
ID: 39727695
In the case of the first example you are trying to pass a template function declaration rather than a template function instance. This isn't allowed. The 2nd version works because you are passing an instantiated object. You can make the first version work by instantiating the template.

std::vector<int> v2;
std::transform(v.begin(), v.end(), std::back_inserter(v2), square<int>);

Notice that I am explicitly instantiating the template function for type int.

Put another way "square" is not a concrete function but "square<int>" is.
squareTempl *is* a concrete object.
0
 

Author Comment

by:Rothbard
ID: 39738440
Thanks! Sorry for the delay in acknowledging your responses.

Merry Christmas :-)
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