CODEWITHSUNDEEP
c++visual-studio-2015decltype
bottaio
bottaio

Reputation: 5093

Deducing type from pointer using decltype

VS2015 is throwing a lot of errors when trying to execute this code:

int a = 5;
int *p = &a;
std::vector<decltype(*p)> v;

However, when I check the type returned by this decltype I get an int!

typeid(decltype(*p)) == typeid(int) // returns true

Can anyone explain it to me? I did the workaround by simply dereferencing pointer and decltyping the value I got. But why isn't it possible to do it by dereferencing pointer directly?

Upvotes: 2

Views: 3644

Answers (2)

R Sahu
R Sahu

Reputation: 206607

As an alternative to the solution proposed by @Brian, you can use:

std::vector<std::remove_reference<decltype(*p)>::type> v;

You can also use:

std::vector<std::remove_pointer<decltype(p)>::type> v;

or

std::vector<std::decay<decltype(*p)>::type> v3;

Upvotes: 11

Brian Bi
Brian Bi

Reputation: 119174

decltype(*p) is int&, not int, and you can't have a vector of references. The unfortunately named typeid doesn't expose this difference since it strips references and cv-qualifiers.

This works though:

 std::vector<decltype(a)> v;

Upvotes: 6

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