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I know that in C++11 we can now use using to write type alias, like typedefs:

typedef int MyInt;

Is, from what I understand, equivalent to:

using MyInt = int;

And that new syntax emerged from the effort to have a way to express "template typedef":

template< class T > using MyType = AnotherType< T, MyAllocatorType >;

But, with the first two non-template examples, are there any other subtle differences in the standard? For example, typedefs do aliasing in a "weak" way. That is it does not create a new type but only a new name (conversions are implicit between those names).

Is it the same with using or does it generate a new type? Are there any differences?

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  • 313
    I personally prefer the new syntax because it is much more similar to regular variable assignment, improving readability. For example, do you prefer typedef void (&MyFunc)(int,int); or using MyFunc = void(int,int); ? Commented May 25, 2012 at 8:08
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    @MatthieuM. those two are different btw. It should be typedef void MyFunc(int,int); (which actually doesn't look as bad), or using MyFunc = void(&)(int,int); Commented May 25, 2012 at 14:28
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    @R.MartinhoFernandes why do you need (&) in using MyFunc = void(&)(int,int); ? does it mean MyFunc is a reference to a function? what if you omit the &? Commented Jul 15, 2015 at 4:50
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    Yes, it's a function reference. It's equivalent to typedef void (&MyFunc)(int,int);. If you omit the & it's equivalent to typedef void MyFunc(int,int); Commented Jul 15, 2015 at 12:24
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    In this episode of CppCast Edouard Alligand claims that using results in faster link times (don't remember which compiler he was talking about, though), because apparently the compiler generates shorter symbol names. I don't know if this holds up to scrutiny, though. Commented Nov 19, 2015 at 15:02

8 Answers 8

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727

They are equivalent, from the standard (emphasis mine) (7.1.3.2):

A typedef-name can also be introduced by an alias-declaration. The identifier following the using keyword becomes a typedef-name and the optional attribute-specifier-seq following the identifier appertains to that typedef-name. It has the same semantics as if it were introduced by the typedef specifier. In particular, it does not define a new type and it shall not appear in the type-id.

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11 Comments

From the answer, using keyword seems to be a superset of typedef. Then, will typdef be deprecated in future ?
Deprecation doesn't necessarily indicate intent to remove--It merely stands as a very strong recommendation to prefer another means.
But then I wonder why they didn't just allow typedef to be templated. I remember having read somewhere that they introduced the using syntax precisely because the typedef semantics didn't work well with templates. Which somehow contradicts the fact that using is defined to have exactly the same semantics.
@celtschk: The reason is talked about in the proposal n1489. A template alias is not an alias for a type, but an alias for a group of templates. To make a distinction between typedef the felt a need for new syntax. Also, keep in mind the OP's question is about the difference between non-template versions.
So why this redundancy whas introduced ? 2 syntaxes for the same purpose. And I don't see typdef being deprecated ever.
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463

They are the same, except that:

The alias declaration is compatible with templates, whereas the C style typedef is not.

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5 Comments

Particularly fond of the simplicity of the answer and pointing out the origin of typeset.
@g24l you mean typedef... probably
What is the difference between C and C++ in typedef if I may ask?
This is not answering the question though. I already know that difference and pointed at in the original post. I was asking only about the case where you don't use templates, is there differences.
@McSinyx, there is no difference.
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The using syntax has an advantage when used within templates. If you need the type abstraction, but also need to keep template parameter to be possible to be specified in future. You should write something like this.

template <typename T> struct whatever {}; template <typename T> struct rebind { typedef whatever<T> type; // to make it possible to substitue the whatever in future. }; rebind<int>::type variable; template <typename U> struct bar { typename rebind<U>::type _var_member; }

But using syntax simplifies this use case.

template <typename T> using my_type = whatever<T>; my_type<int> variable; template <typename U> struct baz { my_type<U> _var_member; }
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I already pointed this in the question though. My question is about if you don't use template is there any difference with typedef. As, for example, when you use 'Foo foo{init_value};' instead of 'Foo foo(init_value)' both are supposed to do the same thing but don't foillow exactly the same rules. So I was wondering if there was a similar hidden difference with using/typedef.
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All standard references below refers to N4659: March 2017 post-Kona working draft/C++17 DIS.

Typedef declarations can, whereas (until C++23) alias declarations cannot, be used as initialization statements

But, with the first two non-template examples, are there any other subtle differences in the standard?

  • Differences in semantics: none.
  • Differences in allowed contexts(+):
    • C++20 and earlier: some.
    • C++23 and onwards: none(++).

(+) Not including the examples of alias templates, which has already been mentioned in the original post.
(++) P2360R0 (Extend init-statement to allow alias-declaration) has been approved by CWG and as of C++23, this inconsistency between typedef declarations and alias declarations will have been removed.

Same semantics

As governed by [dcl.typedef]/2 [extract, emphasis mine]

[dcl.typedef]/2 A typedef-name can also be introduced by an alias-declaration. The identifier following the using keyword becomes a typedef-name and the optional attribute-specifier-seq following the identifier appertains to that typedef-name. Such a typedef-name has the same semantics as if it were introduced by the typedef specifier. [...]

a typedef-name introduced by an alias-declaration has the same semantics as if it were introduced by the typedef declaration.

Subtle difference in allowed contexts

However, this does not imply that the two variations have the same restrictions with regard to the contexts in which they may be used. And indeed, albeit a corner case, a typedef declaration is an init-statement and may thus be used in contexts which allow initialization statements

// C++11 (C++03) (init. statement in for loop iteration statements). for (typedef int Foo; Foo{} != 0;) // ^^^^^^^^^^^^^^^ init-statement { } // C++17 (if and switch initialization statements). if (typedef int Foo; true) // ^^^^^^^^^^^^^^^ init-statement { (void)Foo{}; } switch (typedef int Foo; 0) // ^^^^^^^^^^^^^^^ init-statement { case 0: (void)Foo{}; } // C++20 (range-based for loop initialization statements). std::vector<int> v{1, 2, 3}; for (typedef int Foo; Foo f : v) // ^^^^^^^^^^^^^^^ init-statement { (void)f; } for (typedef struct { int x; int y;} P; auto [x, y] : {P{1, 1}, {1, 2}, {3, 5}}) // ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ init-statement { (void)x; (void)y; }

whereas before C++23 (this answer may have prompted P2360R0 which addressed this niche subtlety in C++23) an alias-declaration is not an init-statement, and thus may not be used in contexts which allows initialization statements

// C++ 11. for (using Foo = int; Foo{} != 0;) {} // ^^^^^^^^^^^^^^^ error: expected expression // C++17 (initialization expressions in switch and if statements). if (using Foo = int; true) { (void)Foo{}; } // ^^^^^^^^^^^^^^^ error: expected expression switch (using Foo = int; 0) { case 0: (void)Foo{}; } // ^^^^^^^^^^^^^^^ error: expected expression // C++20 (range-based for loop initialization statements). std::vector<int> v{1, 2, 3}; for (using Foo = int; Foo f : v) { (void)f; } // ^^^^^^^^^^^^^^^ error: expected expression
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7 Comments

Wow, my intuition was right in the end! There is a difference! Thanks for finding that difference, it's the kind of very very narrow detail that can make a difference in the kind of code I work with (unfortunately XD).
I've never seen this usage of typedef anywhere. Since when is this allowed and what are possible usage patterns of this? Looks kind of horrible to me tbh...
auto [x, y] : {P{1, 1}, {1, 2}, {3, 5}}) { (void)x; (void)y; } What is this called? Where can I find more on this? (Btw, it has one missing open parenthesis.)
@SouravKannanthaB Construction of an std::initializer_list from {…}; skipping the P part starting with the second argument as it’s assumed; structured binding declaration to assign to x and y.
@SouravKannanthaB the opening parenthesis is on the previous line.
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They are essentially the same but using provides alias templates which is quite useful. One good example I could find is as follows:

namespace std { template<typename T> using add_const_t = typename add_const<T>::type; }

So, we can use std::add_const_t<T> instead of typename std::add_const<T>::type

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3 Comments

As far as I know, it's undefined behavior to add anything inside the std namespace
@someonewithpc I was not adding anything, it already exists, I was just showing an example of typename usage. Please check en.cppreference.com/w/cpp/types/add_cv
Is the typename keyword in the RHS optional? I.e., template<typename T> using add_const_t = add_const<T>::type; seems equivalent to the above code I guess. If anyone has advice on this, any comment would be helpful.
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I know the original poster has a great answer, but for anyone stumbling on this thread like I have there's an important note from the proposal that I think adds something of value to the discussion here, particularly to concerns in the comments about if the typedef keyword is going to be marked as deprecated in the future, or removed for being redundant/old:

It has been suggested to (re)use the keyword typedef ... to introduce template aliases:

template<class T> typedef std::vector<T, MyAllocator<T> > Vec;

That notation has the advantage of using a keyword already known to introduce a type alias. However, it also displays several disavantages [sic] among which the confusion of using a keyword known to introduce an alias for a type-name in a context where the alias does not designate a type, but a template; Vec is not an alias for a type, and should not be taken for a typedef-name. The name Vec is a name for the family std::vector<•, MyAllocator<•> > – where the bullet is a placeholder for a type-name.Consequently we do not propose the “typedef” syntax.On the other hand the sentence

template<class T> using Vec = std::vector<T, MyAllocator<T> >;

can be read/interpreted as: from now on, I’ll be using Vec<T> as a synonym for std::vector<T, MyAllocator<T> >. With that reading, the new syntax for aliasing seems reasonably logical.

To me, this implies continued support for the typedef keyword in C++ because it can still make code more readable and understandable.

Updating the using keyword was specifically for templates, and (as was pointed out in the accepted answer) when you are working with non-templates using and typedef are mechanically identical, so the choice is totally up to the programmer on the grounds of readability and communication of intent.

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Both keywords are equivalent, but there are a few caveats. One is that declaring a function pointer with using T = int (*)(int, int); is clearer than with typedef int (*T)(int, int);. Second is that template alias form is not possible with typedef. Third is that exposing C API would require typedef in public headers.

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As of now, C++23 will get typedef and using closer together: P2360 proposes that using constitute an init-statement such as the ones listed in @dfrib's answer as error: expected expression.

However, even with P2360, a typedef cannot be a template.

(Edit [2022-09-14]: This contained incorrect information.)

In total, using is strictly more powerful than typedef, and IMO more readable, too.

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