Re: for loops & 2d array

On Apr 8, 6:47 pm, "}{" <a...@b.com> wrote:
[snip]


[snip]


[snip]

Let me explain what's going on here.
     map[5,2]
is, for build-in arrays, exactly equivalent to
     * ( (map) + (5,2) )
The syntax "x[y]" for built-in arrays is exactly equivalent to "*((x)+
(y))".

The sub-expression "(5,2)" is parsed as two integer literals, the
comma operator acting on them, inside of parenthesis.
The comma operator
-first- evaluates the first part, the integer literal 5,
-then- evaluates the second part, the integer literal 2,
-then- returns the evaluated second part, the integer 2.

     * ( (map) + (5,2) )
is equivalent to
     * ( (map) + (2) )
is equivalent to
     map[2]

Or more simply,
     map[5,2]
is equivalent to
     map[2]
The above is to be crystal clear.

What you want is something more like:
     BYTE (*map)[10] = new BYTE [10] [10];
     //...
     map[5][2] = 0;

Or possibly:
     BYTE *map = new BYTE [10 * 10];
     //...
     map[5*10 + 2] = 0; //stay consistent in your offset calculations

Or if reasonable, avoid the built-in arrays and use std::vector.
     //This initializes x to contain 10 copies of the second argument,
     //and the second argument is a vector containing 10 BYTEs.
     vector<vector<BYTE> > x(10, std::vector<BYTE>(10));

     x[5][2] = 0;

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