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57

answers:

2

I was trying to write an STL method to take the log of a vector:

for_each(vec.begin(),vec.end(),log);

But I get

no matching function for call to ‘for_each(__gnu_cxx::__normal_iterator<double*, std::vector<double, std::allocator<double> > >, __gnu_cxx::__normal_iterator<double*, std::vector<double, std::allocator<double> > >, <unresolved overloaded function type>)’

Which I gather is due to log function's multiple versions. Obviously I can write a simple wrapper around the log function and call it with that. Is there a simpler way to specify which log function I want inline?

+8  A: 

Yes. You can cast the function to the appropriate type:

for_each(vec.begin(),vec.end(),(double(*)(double))log);

Another possibility would be creating your functor that would accept any type:

struct log_f
{
  template <class T> T operator()(const T& t) const { return log(t); }
};

for_each(vec.begin(),vec.end(), log_f());

And, as Billy O'Neal pointed out, you want rather transform than for_each.

jpalecek
+1 and scratched out that part of my answer -- this is the correct answer here.
Billy ONeal
Beautiful, I was originally using transform, but I switched to for_each forgetting that it had different functionality. Thanks, I figured it was some sort of dynamic cast, I just didn't know what the syntax would be. Just so I'm clear that's saying cast log to a function that returns a double and takes a double. Also what does the (*) specify to the compiler in the middle?
Jacob Schlather
This is `static_cast`. the `(*)` in the middle means pointer. ie. `void(int)` is a type of a function that takes an `int` and returns nothing, `void(*)(int)` is a pointer to that function. `void*(int)` would be a function (not a pointer) that takes an `int` and returns a void-pointer.
jpalecek
Ah, makes sense. Thanks again.
Jacob Schlather
It does help to write the static_cast explicitly: `for_each(vec.begin(),vec.end(), static_cast< double(*)(double) >(log));`
MSalters
+2  A: 

I believe std::for_each is looking for a function with a void return type. You're passing a function with a double return type. jpalecek's answer is the correct one, and +1 to him. However, you still have the semantic issue that doing for_each with log doesn't make any sense:

If you want all the members of the vector to be the log of the previous members, i.e.:

//pseudocode
foreach( var x in myvector )
   x = log(x);

Then you don't want for_each, you want transform.

std::transform(vec.begin(), vec.end(), vec.begin(), log);
Billy ONeal