//defines an array of 280 pointers (1120 or 2240 bytes)
int *pointer1 [280];
//defines a pointer (4 or 8 bytes depending on 32/64 bits platform)
int (*pointer2)[280]; //pointer to an array of 280 integers
int (*pointer3)[100][280]; //pointer to an 2D array of 100*280 integers
Using pointer2
or pointer3
produce the same binary except manipulations as ++pointer2
as pointed out by WhozCraig.
I recommend using typedef
(producing same binary code as above pointer3
)
typedef int myType[100][280];
myType *pointer3;
Note: Since C++11, you can also use keyword using
instead of typedef
using myType = int[100][280];
myType *pointer3;
in your example:
myType *pointer; // pointer creation
pointer = &tab1; // assignation
(*pointer)[5][12] = 517; // set (write)
int myint = (*pointer)[5][12]; // get (read)
Note: If the array tab1
is used within a function body => this array will be placed within the call stack memory. But the stack size is limited. Using arrays bigger than the free memory stack produces a stack overflow crash.
The full snippet is online-compilable at gcc.godbolt.org
int main()
{
//defines an array of 280 pointers (1120 or 2240 bytes)
int *pointer1 [280];
static_assert( sizeof(pointer1) == 2240, "" );
//defines a pointer (4 or 8 bytes depending on 32/64 bits platform)
int (*pointer2)[280]; //pointer to an array of 280 integers
int (*pointer3)[100][280]; //pointer to an 2D array of 100*280 integers
static_assert( sizeof(pointer2) == 8, "" );
static_assert( sizeof(pointer3) == 8, "" );
// Use 'typedef' (or 'using' if you use a modern C++ compiler)
typedef int myType[100][280];
//using myType = int[100][280];
int tab1[100][280];
myType *pointer; // pointer creation
pointer = &tab1; // assignation
(*pointer)[5][12] = 517; // set (write)
int myint = (*pointer)[5][12]; // get (read)
return myint;
}
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