Why can’t my parameter be
void example(int Array[][]){ /*statements*/}
Why do I need to specify the column size of the array? Say for example, 3
void example(int Array[][3]){/*statements*/}
My professor said its mandatory, but I was coding before school started and I remembered that there was no syntactical or semantic error when I made this my parameter? Or did I miss something?
When it comes to describing parameters, arrays always decay into pointers to their first element.
When you pass an array declared as
int Array[3]to the functionvoid foo(int array[]), it decays into a pointer to the beginning of the array i.e.int *Array;. Btw, you can describe a parameter asint array[3]orint array[6]or evenint *array– all these will be equivalent and you can pass any integer array without problems.In case of arrays of arrays (2D arrays), it decays to a pointer to its first element as well, which happens to be a single dimensional array i.e. we get
int (*Array)[3].Specifying the size here is important. If it were not mandatory, there won’t be any way for compiler to know how to deal with expression
Array[2][1], for example.To dereference that a compiler needs to compute the offset of the item we need in a contiguous block of memory (
int Array[2][3]is a contiguous block of integers), which should be easy for pointers. Ifais a pointer, thena[N]is expanded asstart_address_in_a + N * size_of_item_being_pointed_by_a. In case of expressionArray[2][1]inside a function (we want to access this element) theArrayis a pointer to a single dimensional array and the same formula applies. The number of bytes in the last square bracket is required to findsize_of_item_being_pointed_by_a. If we had justArray[][]it would be impossible to find it out and hence impossible to dereference an array element we need.Without the size, pointers arithmetics wouldn’t work for arrays of arrays. What address would
Array + 2produce: advance the address inArray2 bytes ahead (wrong) or advance the pointer3* sizeof(int) * 2bytes ahead?