CODEWITHSUNDEEP
c++
user7926026
user7926026

Reputation:

What does it mean that typedef and using are semantically equivalent?

What does it mean the "typedef" and "using" are semantically equivalent? I am mostly focusing on the words "semantically equivalent". I have tried to look "semantically" up, but it did not really mention any synonym or meaning that I could associate with the usage of the word in the programming world.

Upvotes: 2

Views: 123

Answers (2)

Lapshin Dmitry
Lapshin Dmitry

Reputation: 1124

You may think about using as an alias:

using a = b<c>;

can be treated like "As a type, a means b<c>".

typedef does absolutely the same thing, but it comes from C language and uses the same syntax as variable declarations, and it is hard to read them when used with multiple-layered templates, arrays, cv-classifiers, function types etc:

typedef int const* (*func_t)(std::vector<int> const&);
// The same:
using func_t = int const* (*)(std::vector<int> const&);

For me, using is a lot better because you can easily see the name.

Another difference is that you can use using with template arguments. For some time, I used this in my hacking code:

template<typename T>
using set_it = std::set<T>::iterator;

And every time I used set_it<int> as a type I got a std::set<int>::iterator. You cannot do this with typedef.

Upvotes: 1

PointerToConstantChar
PointerToConstantChar

Reputation: 146

It just means they do the same thing, Even if they are slightly different in syntax*. For example the following:

typedef int INTEGER;

Can be written with the using syntax as follows:

using INTEGER = int;

*The using syntax works also with templates where typedef doesn't, But for non-templates they are equivalent.

Upvotes: 1

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