CODEWITHSUNDEEP
c++arraysmultidimensional-array
user34537
user34537

Reputation:

assign to int[][4] in C/C++

How do I assign to a multi dimensional array without casting [][] to **? I have an example below. What I found was How do I declare a 2d array in C++ using new? but that uses int** rather then new int[][4].

#include<cassert>

int a[8][4];
int*b = &a[0][0];
int*c = &a[2][0];
int*d = &a[0][2];

int main() {
    //shows right side is closer
    assert(d-b==2);
    assert(c-b==8); 
    auto aa = new int[][4];
    //set the right side, but is a syntax error
    //aa[][0] = new int[8];
    //type error
    aa[0] = new int[8];
}

Upvotes: 0

Views: 157

Answers (3)

alecbz
alecbz

Reputation: 6488

edit: https://stackoverflow.com/a/18723991/598940 indicates a = new int[][8] is actually not legal.

From your comments, are you asking why a = new int[][8] is legal but a[][4] is not?

int[] is a type expression representing the same as the type int*. It exists (I believe) to allow for more descriptive type declarations (perhaps among other things)

void foo(int* x);   // is x an array of ints? or just a pointer to a single one?
void foo(int[] x);  // ah, ok -- x is intended to represent an array of ints

The value expression a[], on the other hand, has no meaning.

Upvotes: 0

James Kanze
James Kanze

Reputation: 153909

You should be getting an error on the line with the new int[][4]. There must be an expression inside the []. Otherwise, how can the compiler know how much to allocate. (Some quick checks with VC++, which erroneously does accept this expression, shows that it treats it as the equivalent of new int[0][4], and allocates 0 bytes. It doesn't compile with g++.)

And of course, you confuse things additionally by abusing auto. The one reason never to use auto is that you don't know the actual type, and auto means you don't have to know it (until you want to use the variable, of course). In this case, you should definitely not use auto, because the type is a bit special, and you want your readers to know it:

int (*aa)[4] = new int[8][4];

Once you see this, it should be obvious that aa[0] has type int[4], and there's no way you can assign a pointer to it, since it isn't a pointer.

Of course, in C++, you'd never write anything like this anyway. You'd define a Matrix2D class, and use it. (And it would probably use a std::vector<int> to get the memory, and calculate the single index from the two you give. Internally, of course, so you don't have to think about it.) There is, in fact, never a case for using array new.

Upvotes: 1

Sergey L.
Sergey L.

Reputation: 22542

Try

int (*array)[4] = (int (*)[4])malloc(sizeof(int) * 8 * 4);

now you can access up to

array[7][3];

Upvotes: 1

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