是否可以将指针和数组作为函数参数传递？

Question

I have two integer arrays created at runtime (size depends on the program input). At some point I need to update the contents of an array with the contents of the other doing some calculations.

First I thought about passing those arrays as parameters to a function because I didn't find a way to return functions in C (don't think it's possible). After realizing that was a bad idea since parameters are not really modifiable as they're copied to the stack I resorted to change to array pointers instead.

While the function is still empty, this is the code I have:

1st take (code compiles, no errors):

// Elements is just to be able to iterate through their contents (same for both):
void do_stuff(int first[], int second[], int elements) {}

// Call to the function:
do_stuff(first, second, elements);

2nd take, attempt to translate to pointers to be able to modify the arrays in place:

void do_stuff(int *first[], int *second[], int elements) {}

// Call to the function:
do_stuff(&first, &second, elements);

This code lead to some rightful compile time errors, because apparently what I thought to be pointer to arrays were arrays of pointers.

3rd take, what I think it'd be the right syntax:

void do_stuff(int (*first)[], int (*second)[], int elements) {}

// Call to the function:
do_stuff(&first, &second, elements);

Still this code produces compile time errors when trying to access the elements of the arrays (eg *first[0] ):

error: invalid use of array with unspecified bounds

So my question is regarding the possibility of using an array pointer as a parameter of a function, is it possible? If so, how could it be done?

Anyway, if you think of a better way to update the first array after performing calculations involving the contents of the second please comment about it.

Answer 1

An array decays to a pointer to the data allocated for the array. Arrays are not copied to the stack when passing to functions. Thus, you needn't pass a pointer to the array. So, the below should function fine.

 // Elements is just to be able to iterate through their contents (same for both): void do_stuff(int first[], int second[], int elements) {} // Call to the function: do_stuff(first, second, elements); 

The cause of your errors on your second attempt are because int *first[] (and the others like it) are actually of the type array of pointer to int .

The cause of your third errors are because *first[N] is actually *(first[N]) , which cannot be done with ease. Array access is really a facade over pointer arithmetic, *(first + sizeof first[0] * N) ; however, you have an incomplete element type here -- you need to specify the size of the array, otherwise sizeof first[0] is unknown.

Answer 2

Your first attempt is correct. When passing an array as a parameter in C, a pointer to the first element is actually passed, not a copy of the array. So you can write either

void do_stuff(int first[], int second[], int elements) {}

like you had, or

void do_stuff(int *first, int *second, int elements) {}

Answer 3

In C arrays automatically decay to pointers to data, So, you can just pass the arrays and their lengths and get the desired result.

My suggestion is something like this:

void dostuff(int *first, int firstlen, int *second, int secondlen, int elements)

Function call should be:

 do_stuff(first, firstlen, second, secondlen, elements);

I am not very clear from your question, why you need elements . But, you must pass array lengths as arrays automatically decays to pointers when passed to a function, but, in the called function, there is no way to determine their size.