Creating array of 2d arrays

Question

I have a number of arrays

double foo[][2] = { {1.0,3.0}, {2.6,10.0}, {0.0,0.0} };
double bar[][2] = { {1.4,3.2}, {2.1,9.9}, {2.1,9.9}, {2.1,9.9}, {0.0,0.0} };

So these are both of type:

double (*)[2]

I want to make an array of these so I need to declare an array of type pointer to array[2]

double ((*array)[2])[] = {foo, bar};

This syntax doesn't work - is it possible to do this in C.

I also tried to create a type but this didn't work either:

typedef double *mytype[2] ;
mytype array[] = {foo, bar};

Marcelo Cantos · Accepted Answer · 2010-09-03 13:33:48Z

3

The two arrays are not of type double(*)[2]. foo is of type double [3][2] and bar is of type double [5][2]. The outer array size is implied, not absent.

To create an array of these, use them as pointers, thus:

typedef double (*mytype)[2];
mytype array[] = { foo, bar };

edited Sep 3, 2010 at 13:33

answered Sep 3, 2010 at 13:25

Marcelo Cantos

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3 Comments

Callum Over a year ago

The outer array size can be implied. This isn't the issue I see it is creating an array with the 1st element foo and the 2nd bar.

Marcelo Cantos Over a year ago

The two types are different, so you can't have an array that holds both. So the fact that the implied sizes are different is quite significant. Anyway, the code I presented compiles and runs.

Callum Over a year ago

It does work. It is not purely pointer arithmetic that is being used to find the value of array[0][2][0]. array[0] decays into foo : so foo[2][0].

caf · Accepted Answer · 2010-09-04 09:43:17Z

2

The correct syntax to do it without the typedef is:

double (*array[])[2] = {foo, bar};

answered Sep 4, 2010 at 9:43

caf

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Comments

Sadeq · Accepted Answer · 2010-09-03 13:29:26Z

0

double ((*array)[2])[] = {foo, bar};

This don't work, because the memory location of bar is not relative to the memory location of foo, and the type of foo and bar isn't equal with the type of elements of double (*)[2] which have to be double type.

answered Sep 3, 2010 at 13:29

Sadeq

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1 Comment

Callum Over a year ago

foo and bar decay into pointers - the issue is how to create an array of pointers to arrays.

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