Is it possible to cast an array of one type to an array of another type whose size is different?

Question

For example, can I take

int array[12];

and cast it to a char[48] simply by casting the pointer, and given the assumption that int is 4 bytes on my machine? What would be the proper syntax for this, and would it apply generally?

I understand that the size of the new array wouldn't be explicit, i.e. I'd have to do the division myself, again, knowing that on my machine int is 4 bytes.

I wanted to serialize an integer array to a string (or c-string) without using any library like boost::serialization. It's for a coding exercise. I know I can do it the "naive" way, but wondered about this. — Andrew Cheong
– Andrew Cheong, Commented Feb 24, 2017 at 21:42
Casting pointers may or may not cause any problems. That doesn't indicate anything about whether there will be problems downstream. That depends on how you use the pointers — R Sahu
– R Sahu, Commented Feb 24, 2017 at 21:46
Do you really want to cast int array[12] (48 bytes on your platform) to char array[3] (3 bytes)? Or did you mix up the array bounds? — IInspectable
– IInspectable, Commented Feb 24, 2017 at 23:16
@IInspectable - I just did the math backwards in my example. Correcting, thanks. — Andrew Cheong
– Andrew Cheong, Commented Feb 25, 2017 at 23:51

Community · Accepted Answer · 2020-06-20 09:12:55Z

6

The proper C++ way is with reinterpret_cast :

int array[12];
char* pChar = reinterpret_cast<char*>(array);

You should note that sizeof(array) will be 12*sizeof(int) which is equal to 12 * (sizeof(int) / sizeof(char)), which in most (if not any) machine is larger than sizeof(char[3]) since sizeof(char) is usually a quarter of sizeof(int).

So int array[12]; can be interpeted as char array[12*4];

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Feb 24, 2017 at 21:45

ZivS

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1 Comment

IInspectable Over a year ago

An int is larger than a char on any platform (int is at least 16 bits wide).

François Andrieux · Accepted Answer · 2017-02-24 21:46:29Z

1

You can cast most pointer types to most other pointer types using reinterpret_cast.

Usage to : auto ptr = reinterpret_cast<char*>(array);.

Rather than rely on the fact that int is 4 bytes on your system, you can use sizeof(int) as a constant instead. The wording of your question makes it seem like you may be confused about the "size" of the array when cast to char*. int is sizeof(int) times larger than char. If sizeof(int) == 4 then your ints take up 4 * 12 = 48 bytes in your array, not 3.

answered Feb 24, 2017 at 21:46

François Andrieux

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Comments

user2100815 · Accepted Answer · 2017-02-24 21:42:56Z

0

 int array[12];
 char * p = (char *) array;
 // do what you want with p

Of course, p doesn't have the type char[3], but this probably doesn't matter to you.

answered Feb 24, 2017 at 21:42

user2100815

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