CODEWITHSUNDEEP
c++
Oliver Rasmussen
Oliver Rasmussen

Reputation: 17

Understanding parameter (T *const *)

I'm trying to create a pointer to a vector of floats that I have, so that I can pass it to a generic function that looks something like this:

foo<Type>(Type *const *, size_t)

However, as I'm somewhat new to c++, I'm having difficulties understanding how the syntax in the first parameter should be interpreted, in particular the " *const * ". What exactly would I need to put as an argument?

Any help would be appreciated!

Upvotes: 0

Views: 61

Answers (1)

Remy Lebeau
Remy Lebeau

Reputation: 597156

const applies to the thing on its left, unless there is nothing there, then it applies to the thing on its right instead.

So, in Type *const *, the const applies to the 1st *. So this would mean that the parameter is a non-const pointer (the 2nd *) to a const pointer (the 1st *) to a non-const Type instance. Which would be written like this when calling foo() with a vector of floats:

vector<float> vec;
// populate vec as needed...
float* const ptr = vec.data(); // or: ... = &vec[0];
foo<float>(&ptr, vec.size());

However, depending on what foo() actually does, you probably don't need that extra level of indirection and can remove one of the pointers:

foo<Type>(const Type *, size_t);

vector<float> vec;
// populate vec as needed...
foo<float>(vec.data()/* or: &vec[0] */, vec.size());

Upvotes: 3

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