How to convert a function that returns unique_ptr into raw pointer?

Question

Suppose there is a function in a legacy library that I want to use that requires a function pointer as an input

void LegacyFunction(int* (*func)(float, float), int a, float b);

but the problem is it is expecting the return value of the function to be an int raw pointer instead of an int unique_ptr, meaning I can only get it to compile if my function looks like

int* MyFunc(float a, float b);

in other words, if I modify MyFunc into

std::unique_ptr<int> MyFunc(float a, float b);

and pass it into the legacy library function like

LegacyFunction(MyFunc, 1, 2.0f);

there will be a compilation error. I know if the function is taking an usual int pointer there can be some workaround using the get() function like

std::unique_ptr<int> a;
LegacyFunctionRawPointer(a.get(), 1, 2.0f);

Is there a similar workaround for function pointer input? It will be a travesty if I have to replace the unique_ptr for MyFunc with raw pointer.

Is MyFunc a runtime value or is it known at the compile time? — krzaq
– krzaq, Commented Nov 7, 2016 at 22:58
Who owns the pointer? Is legacy library the owner? Does legacy library destroy the pointer? — Robinson
– Robinson, Commented Nov 7, 2016 at 22:58
A simple lambda function wrapper seems like it would work, but you'd have memory leaks if the legacy didn't free the memory returned in the same way it was allocated. — ShadowRanger
– ShadowRanger, Commented Nov 7, 2016 at 22:59

ShadowRanger · Accepted Answer · 2016-11-07 23:03:14Z

5

A lambda function should work, e.g.:

LegacyFunction([](float a, float b) { return MyFunc(a, b).release(); }, 1, 2.0f);

but only if LegacyFunction will naturally delete/free the memory that MyFunc allocated with the appropriate operation (so if MyFunc used new int, LegacyFunction must use delete retval;; if it used new int[x], it must use delete[] retval;).

The release method gets the contained pointer and relinquishes ownership, so you've explicitly given up unique_ptr's protections; not a great idea if you can avoid it.

answered Nov 7, 2016 at 23:03

ShadowRanger

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