In C++ 03, how to refer to the address of an instiantiated template function?

Question

Assuming that & is not overloaded. How to obtain the address of an instantiated template function, such as std::sort<int*>? The following won't compile on some compilers:

#include <algorithm>
int main()
{
    &std::sort<int*>;
}

ON MSVC v19.21, it reports: https://godbolt.org/z/gpZCdn

error C2568: 'identifier': unable to resolve function overload

Specifying the template arguments of a standard algorithm is UB IIRC. — L. F., Jul 18 '19 at 04:57
Possible duplicate of [How do I specify a pointer to an overloaded function?](https://stackoverflow.com/questions/2942426/how-do-i-specify-a-pointer-to-an-overloaded-function) — Alan Birtles, Jul 18 '19 at 05:03
@JeJo Thanks, but the question focuses on a more generic situation. The code snippet is just used as an example. — xuhdev, Jul 18 '19 at 05:12

R Sahu · Accepted Answer · 2019-07-18T05:18:42.877

3

You can use

&std::sort<int*>;

&std::sort<int> does not work since the type needs to be dereferenceable.

The ambiguity can be resolved by performing an explicit cast.

static_cast<void (*)(int*, int*)>(&std::sort<int*>);

edited Jul 18 '19 at 05:18

answered Jul 18 '19 at 04:53

R Sahu

204,454
14
159
270

no, `std::sort` will indeed not compile but the issue the OP is having is that the overload is ambiguous – Alan Birtles Jul 18 '19 at 05:05
I edited the question so that the ambiguity in the question itself (around `std::sort`) is clarified. – xuhdev Jul 18 '19 at 05:07
@xuhdev, interesting. It works with g++ 5.4 at the same site. – R Sahu Jul 18 '19 at 05:09
@RSahu I know it works with gcc, but it doesn't seem to be supported by all compilers, even if they fully support C++ 03. clang 3.2 also won't compile this example. – xuhdev Jul 18 '19 at 05:15
@xuhdev, did you try the `static_cast`? – R Sahu Jul 18 '19 at 05:16

robthebloke · Answer 2 · 2019-07-18T05:17:28.920

3

Sort doesn't depend on a type, it depends on an iterator type.
There are multiple overloads of std::sort, so you need to help the compiler choose which one via the function prototype.
No need to use & to get the address - the name is the address.

Example:

void (*func_ptr)(std::vector<int>::iterator, std::vector<int>::iterator) =
     std::sort< std::vector<int>::iterator >;

std::vector<int> values;
for(int i = 99; i > 0; --i) 
  values.push_back(i);

func_ptr(values.begin(), values.end());

And if you really want int* as your iterator type

void (*func_ptr)(int*, int*) = std::sort<int*>;

edited Jul 18 '19 at 05:17

answered Jul 18 '19 at 04:57

robthebloke

9,331
9
12

std::sort returns void. You are missing void at the start of the function pointer declaration. See the void in my example above. Just keep in mind that the sort function will work only on the iterator type you have specified. Your simplified example will only work on raw C arrays, and not with lists/vectors/maps etc. – robthebloke Jul 18 '19 at 05:15

In C++ 03, how to refer to the address of an instiantiated template function?

2 Answers2