For some reason this is giving me more trouble than i thought...
int *myArray[3];
myArray = new int[mySize];
does not work...
I've used a typedef before in a similar manner and it worked perfectly, but this time i dont want to create the typedef
One might be tempted to do this:
::std::vector<int[3]> myArray;
Because vector is so nice for dynamically sized arrays. Unfortunately, while that declaration works, the resulting vector is unusable.
This will be just as efficient, if you have ::std::array (a C++11 feature) and it will actually work:
::std::vector< ::std::array<int, 3> > myArray;
If you can do this, I would highly recommend it. vector is much nicer and safer to deal with than an array you have to allocate yourself with new.
Otherwise, try this:
typedef int inner_array_t[3];
inner_array_t *myArray = new inner_array_t[mySize];
And since you don't want to use a typedef for some odd reason, you can unwrap it like so:
int (*myArray)[3] = new int[mySize][3];
std::vector<int[3]> myArray; actually works, because C style arrays do not fulfill the requirements of a vector. What would you push_back into the vector? Something like int a[3]; myArray.push_back(a); doesn't work.
std::vector<std::array<int, 3>>.int (*myArray)[3] = new int[mySize][3];
?
int *myArray[3];
This means "myArray shall be an array of three pointers-to-int".
You presumably wanted "myArray shall be a pointer-to-(array of three ints)". That is spelled int (*myArray)[3].
This sort of thing is much easier with typedefs.
typedef int datum[3];
datum* myArray = new datum[mySize]; // no fuss, no muss.
But seriously, just use std::vector. And make an actual struct for your group-of-three-integers. Or, if it really should behave like an array, use boost::array so that you at least get the behaviour of a first-class type.
myArray[0]=new int[mySize];?