How to assign a multi-dimensional array to a temporary variable?

Question

I want to assign a statically allocated, multi-dimensional array to a temporary variable. Consider the following example:

void foo(int b[3][2])
{
    b[1][1] = 1; // no segmentation fault
}

int main()
{
    int a[3][2] = {{1, 2}, {11, 12}, {21, 22}};

    foo(a);

    int** c;
    c = (int**)&a;
    c[1][1] = 1; // segmentation fault on execution

    int* d[3];
    d[0] = (int*)&(a[0]);
    d[1] = (int*)&(a[1]);
    d[2] = (int*)&(a[2]);
    d[1][1] = 1; // no segmentation fault

    return 0;
}

Basically I want to do what the compiler does with the parameter b of foo(). But the only working solution I could come up with is d. Is there no less complicated way?

It's strange how many people think a 2D array somehow casts directly to a pointer to pointers. — tenfour
– tenfour, Commented Nov 21, 2011 at 14:36

Robᵩ · Accepted Answer · 2011-11-21 14:39:58Z

12

cdecl (man page) is your friend:

cdecl> explain int b[3][2]
declare b as array 3 of array 2 of int
cdecl> declare b as pointer to array 2 of int
int (*b)[2]

So, try this:

void foo(int b[3][2])
{
    b[1][1] = 1; // no segmentation fault
}

int main()
{
    int a[3][2] = {{1, 2}, {11, 12}, {21, 22}};

    foo(a);

    int (*b)[2] = a;

    b[1][1] = 1;

    return 0;
}

edited Nov 21, 2011 at 14:39

answered Nov 21, 2011 at 14:34

Robᵩ

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2 Comments

sehe Over a year ago

huh - that's new. where/what is cdecl?

Kiril Kirov Over a year ago

OMG cdecl :D:D:D I have never seen this :D There's online version: @sehe - cdecl.org

Sarfaraz Nawaz · Accepted Answer · 2011-11-21 15:10:52Z

9

int[3][2] and int** are incompatible types. You cannot cast one to another.

Try this:

int (*c)[2];
c = a; //no need to cast 
c[1][1] = 1; //ok

Or you could do this (declaration as well as initialization):

int (*c)[2] = a; //no need to cast 
c[1][1] = 1; //ok

Thumb of rule:

Don't use c-style cast in C++. Use C++-style cast. Had you used C++-style cast, the compiler would have told you the problem much before (ideone) (no need to run the code to see the problem):
```
prog.cpp:5: error: invalid static_cast from type ‘int (*)[3][2]’ to type ‘int**’
```
But C-style cast compiles it fine (ideone), as you already know.
And whenever you use cast, even C++-style cast, your first doubt should be the cast itself if the program doesn't work as expected.

edited Nov 21, 2011 at 15:10

answered Nov 21, 2011 at 14:34

Sarfaraz Nawaz

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Comments

sehe · Accepted Answer · 2011-11-21 14:47:25Z

6

As you are probably aware now, from the other answers, the type of a is not actually equivalent to int** - it jsut decays to that (when returned/passed by value).

int (*b)[2] = a; // would solve that

There is a more C++ way:

typedef std::array<std::array<int, 2>, 3> M23;

void foo(M23& b)
{
    b[1][1] = 1; 
}

int main()
{
    M23 a = {{1, 2}, {11, 12}, {21, 22}};

    foo(a);

    M23 d = a;
    d[1][1] = 1;
}

edited Nov 21, 2011 at 14:47

answered Nov 21, 2011 at 14:36

sehe

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1 Comment

Sarfaraz Nawaz Over a year ago

@sehe: This is good, but you should also answer the question, and explain the problem with OP's code. Once you explained that, you can suggest better alternative solution.

Some programmer dude · Accepted Answer · 2011-11-21 14:35:00Z

3

If you are using a pretty modern compiler that supports enough parts of the C++11 standard, you can use auto:

int a[3][2] = ...;
auto &b = a;
b[1][1] = 1;  // a[1][1] will be set

Of course, both a and b has to be defined in the same scope for it to work. You can't have an auto parameter in a function for example (that's what templates are for.)

answered Nov 21, 2011 at 14:35

Some programmer dude

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Comments

Basile Starynkevitch · Accepted Answer · 2011-11-21 14:36:55Z

2

Don't cast explicitly, so try writing

 c = &a;

Then the GCC compiler (using gcc -Wall -g bidim.c -o bidim to compile) gives you the correct warning:

bidim.c:13:7: warning: assignment from incompatible pointer type [enabled by default]

And then you should realize that a 2D matrix is not implemented as an array of pointers to 1D arrays.

answered Nov 21, 2011 at 14:36

Basile Starynkevitch

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Comments

Michael Krelin - hacker · Accepted Answer · 2011-11-21 14:37:19Z

0

The first thing that comes into my mind is using a typedef and reference like

typedef int thing_t[3][2];
thing_t& e = a;
e[1][1] = 1;

With pointers

int (*f)[2] = a;
f[1][1] = 1;

Another possibility is to encapsulate it in struct.

answered Nov 21, 2011 at 14:37

Michael Krelin - hacker

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