When defining a function like this:
void myFunction(arguments){
// some instructions
}
What is the deference between using type name[] and type *name as the function's argument.
When using type name[] and calling the function does it make a copy to the name array or just point to the passed array.
Is the name array's subscript needs to be set. If not what's the difference.
There is no difference to my knowledge which is actually a problem in this situation:
int func(int a[20]) {
sizeof(a); // equals sizeof(int*)!!
}
Therefore I always prefer the pointer form, as the array form can cause subtle confusion.
Either way only a pointer is passed.
void f(int array[]);
is synonymous to passing a const pointer
void f( int * const array);
They become non-synonymous when passing two dimensional arrays with bounds
void f(int array[][n]);
is synonymous to
void f(int (* const array)[n]);
But different to
void f(int **array);
int array[][n] is an array of pointers to int where int **array is a pointer to pointer to int. Is that right?
int array[][n] is an array of/a pointer to an undefined amount of arrays of n ints (single block of memory). While int ** array is an array of pointers to int (each of which can point to it's own separate block of memory holding an undefined amount of ints).array is just the name otherwise you means something else
int array[] parameter is not adjusted to int * const array. It is just int *array; there is no const. If the parameter were int array[const], then it would be adjusted to int * const array. Per C 2011 (N1570) 6.7.6.3 7: “A declaration of a parameter as ‘‘array of type’’ shall be adjusted to ‘‘qualified pointer to type’’, where the type qualifiers (if any) are those specified within the [ and ] of the array type derivation.”
array is int *In C++ it's the same. Different is in declaration;
char *s = "String"; // allocated and fill
char str[80]; // allocated memory
char *p1; // not allocated memory
p1 = str;
