CODEWITHSUNDEEP
c++c++11typedef
Somethone
Somethone

Reputation: 269

Using c++ typedef/using type alias

I am reading C++ primer book and completely don't understand one line:

 using int_array = int[4]; 
 typedef int int_array[4]; // This line
 for (int_array *p = ia; p != ia + 3; ++p) {
      for (int *q = *p; q != *p + 4; ++q)
          cout << *q << ' '; cout << endl;
 }

Ok typedef is same as using. Does it mean int[4][4] is now int and how to understand that? And what type is int_array in for loop?

Thanks

Upvotes: 26

Views: 27287

Answers (3)

wasmup
wasmup

Reputation: 16233

Both type aliases are the same:

Type alias, alias template (since C++11):
Type alias is a name that refers to a previously defined type (similar to typedef):

using identifier attr(optional) = type-id ;

so you may use:

typedef int int_array[4];

or you may just use (it is the same as above):

using int_array = int[4];

When you need to address the memory with 4*sizeof(int) steps, e.g. if the system int size is 4 bytes, then the memory step size is 4*4=16 bytes. even you may use int_array *p; in this case ++p advances p by one memory step e.g. 16 bytes. see:

1- working sample with using int_array = int[4];:

#include <iostream>
using std::cout; using std::endl;

int main()
{
    int ia[3][4] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 };

    // a range for to manage the iteration
    // use type alias
    using int_array = int[4];
    for (int_array& p : ia)
        for (int q : p)
            cout << q << " ";
    cout << endl;

    // ordinary for loop using subscripts
    for (size_t i = 0; i != 3; ++i)
        for (size_t j = 0; j != 4; ++j)
            cout << ia[i][j] << " ";
    cout << endl;

    // using pointers.
    // use type alias
    for (int_array* p = ia; p != ia + 3; ++p)
        for (int *q = *p; q != *p + 4; ++q)
            cout << *q << " ";
    cout << endl;

    return 0;
}

output 1:

0 1 2 3 4 5 6 7 8 9 10 11
0 1 2 3 4 5 6 7 8 9 10 11
0 1 2 3 4 5 6 7 8 9 10 11

2- working sample using typedef int int_array[4];:

#include <iostream>
using std::cout; using std::endl;

int main()
{
    int ia[3][4] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 };

    // a range for to manage the iteration
    // use type alias
    typedef int int_array[4];
    for (int_array& p : ia)
        for (int q : p)
            cout << q << " ";
    cout << endl;

    // ordinary for loop using subscripts
    for (size_t i = 0; i != 3; ++i)
        for (size_t j = 0; j != 4; ++j)
            cout << ia[i][j] << " ";
    cout << endl;

    // using pointers.
    // use type alias
    for (int_array* p = ia; p != ia + 3; ++p)
        for (int *q = *p; q != *p + 4; ++q)
            cout << *q << " ";
    cout << endl;

    return 0;
}

output 2(same):

0 1 2 3 4 5 6 7 8 9 10 11
0 1 2 3 4 5 6 7 8 9 10 11
0 1 2 3 4 5 6 7 8 9 10 11

Ref: https://github.com/Mooophy/Cpp-Primer/blob/master/ch03/ex3_44.cpp
Note: use -std=c++11 for compile/link.

Upvotes: 1

user4581301
user4581301

Reputation: 33931

TL;DR

Both are doing the exact same thing: Defining int_array as an alias of an array of 4 ints

Making Sense of the Syntax

using has a nice A = B notation that is generally much easier to understand.

using alias = type;

typedef's notation is not quite backward. For a simple typedef

typedef type alias;

but more complicated typedefs tend to sprawl. I suspect the syntax was modeled after how one would define a variable, but I can't find where I packed my copy of the old K&R C programming book and can't look that up at the moment.

int int_array[4];

would define int_array to be an array of 4 ints. Slapping typedef on the front

typedef int int_array[4];

makes int_array a type alias instead of a variable.

Another example,

int * intp;

Defines intp to be a pointer to an int.

typedef int * intp;

Defines intp to be an alias to the type pointer to an int.

This gets ugly with more complicated data types as the name of the typedefed alias may be buried somewhere in the middle of the definition. A typedefed function pointer for example:

typedef void (*funcp)(param_t param1, param_t param2, ...);

vs using

using funcp = void (*)(param_t param1, param_t param2, ...);

Making a 2D Array

If you want a 2D array you could

using int_array2D = int[4][4];

or you could define an array of int_array

using int_array2D = int_array[4];

And of course that means you can

using int_array3D = int_array2D[4];

and keep on going until the cows come home or you've packed on so many dimensions that The Doctor's brain melts.

Upvotes: 24

plats1
plats1

Reputation: 930

This line does nothing as it is redundant

The lines

using int_array = int[4];

and

typedef int int_array[4];

do the same thing. See reference for using and typedef. You can leave one or the other out and the behaviour is the same. It is not an error to have two different declarations, as they are not conflicting (they do the exact same thing).

The first way of doing it (using the using keyword) was introduced with C++11 and is in my opinion easier to read, so I prefer it over the typedef version.

Upvotes: 6

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