weird behaviour of "size of"

[ünloco]

hello
i have this array

[RF_Developer]

Hmm... in C nums[0] is a pointer. Normally sizeof(a pointer) returns the size of the pointer, i.e. 2 in CCS generally. Using sizeof() with an array is a special case, and should the size of the array, not the the size of the pointer. C implements arrays with pointers remember and C arrays can be indexed or accessed through pointers interchangably. Note, if you had:

n1 = 2;

int *P = nums[2]; // P is now interchangable with nums[2].

then the following should happen:

width = sizeof nums[2]; //gives width = 6 and that's ok
width = sizeof nums[n1]; //gives width = 2 !!!
width = sizeof(P); // should be 2, i.e. sizeof(int *)

So, for some reason, CCS is interpreting sizeof(nums[2]) as the special case - tick VG - but sizeof(nums[n2]) as the normal, i.e. pointer case - not so good, must do better. The use of the variable is breaking the special case of sizeof for arrays... However the two are evaluated at very different times. sizeof(nums[2]) is in this case known at compile time, whereas sizeof(nums[n1]) has to be evaluated at run time. If nums[] was a ragged array, i.e. its elements not defined at compile time then sizeof(nums[2]) would also not be known at compile time. This makes sizeof() more tricky for compiler writers than we might at first expect.

RF Developer

[Ttelmah]

Sizeof, is computed at compile time, so trying to access an element indexed by a variable simply won't work. You get '2', because the compiler decides you want the storage size of the pointer to the row represented, and this (of course) is 2.

I see RF_Developer has pointed this out as well, while I'm typing.

Best Wishes

[ünloco]

thank you guys !
now it's clearer as to why
i'll avoid such use
greetz
_________________
for(;;);