Re: Some doubts on variable-length array

On Fri, 31 Mar 2006 02:00:45 -0800, lovecreatesbeauty wrote:

I can understand the behaviour of these statements marked `/* A. don't
understand */' in following code, but do not understand the statements
marked `/* B. understand */' .

Did you mean it that way round? I've commented on both just in case.

#include <stdio.h>

void foo(int size_x, int size_y, int tab[size_x][size_y])
{
printf("tab[1][1] == %d\n", tab[1][1]);

prints the 2nd element of the 2nd element of tab.

}

int main()
{
/* int tab[9] = {0, 1, 2, 3, 4, 5, 6, 7, 8}; */
int tab[3][3] = {0, 1, 2, 3, 4, 5, 6, 7, 8};

My compiler (gcc 4.0.1) warns me I should write:

int tab[3][3] = {{0, 1, 2}, {3, 4, 5}, {6, 7, 8}};

although it accepts the flat list as well.

foo(2, 2, tab); /* A. don't understand */

This calls "lies" to foo telling it that the array is 2x2 so foo sees it
as if it were "int tab[2][2] = {{0, 1}, {2, 3}};". Element [1][1] is 3
and this is what is printed.

printf("%i\n", tab[2][2]); /* B. understand */

The declaration of tab is in scope here so there should be no confusion.
The third element of the third element of tab is 8.

printf("%i\n", *(*(tab+2)+2)); /* B. understand */

This is another way to write the same thing. The array name tab is
treated as a pointer to its first element (an array of three ints).
Adding 2 to this pointer (tab + 2) moves it by two "strides" to make a
pointer that points to the third element (another array of three ints).
*(tab + 2) is this array, but again, the array is treated as a pointer to
its first element (an int). Adding 2 to that gives a pointer that points
to the number 8. The final * de-references that pointer to give the value
8.

printf("\n\n");

foo(3, 3, tab); /* A. don't understand */

This call does not "lie". So inside foo it is seen as it was defined and
element [1][1] is the number 4.

printf("%i\n", tab[3][3]); /* B. understand */
printf("%i\n", *(*(tab+3)+3)); /* B. understand */

These two are exactly as above, but the array is being index out of bounds
so you get undefined behaviour (probably a segmentation fault, or maybe
just garbage being printed).

printf("\n\n");

return 0;
}

--
Ben.
.

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