Why does this work:
void SomeFunction(int SomeArray[][30]);
but this doesn't?
void SomeFunction(int SomeArray[30][]);
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2 Answers
Because, when passing an array as an argument the first [] is optional, however the 2nd onwards parameters are mandatory. That is the convention of the language grammar.
Moreover, you are not actually passing the array, but a pointer to the array of elements [30]. For better explanation, look at following:
T a1[10], a2[10][20];
T *p1; // pointer to an 'int'
T (*p2)[20]; // pointer to an 'int[20]'
p1 = a1; // a1 decays to int[], so can be pointed by p1
p2 = a2; // a2 decays to int[][20], so can be pointed by p2
Also remember that, int[] is another form of int* and int[][20] is another form of int (*)[20]
2 Comments
Benjamin Lindley
"int[] is another form of int* and int[][20] is another form of int (*)[20]" -- Only in parameter lists. In normal declarations, int[] is an array of however many ints are in its initialization list, likewise with int[][20].
iammilind
@BenjaminLindley, you are correct. My answer was in the context of the argument passing only.
Intuitively, because the compiler cannot compute a constant size for the elements of the formal in the second declaration. Each element there has type int[] which has no known size at compile time.
Formally, because the standard C++ specification disallow that syntax!
You might want to use std::array or std::vector templates of C++11.
answered Apr 28, 2012 at 6:10
Basile Starynkevitch
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