Passing an array by pointer

Question

How do i get the size of array in void Func(int (*a)[5]) for loop condition and print elements?

void Func(int (*a)[5])
{
   for (int i = 0; i < 5; i++)
      cout << a[i]; // doesn't work
}

int main()
{
   int a[] = {1, 2, 3, 4, 5};
   Func(&a);
}

@JT (a very annoying tag, BTW) Not generally possible would be more accurate :-) — , May 12 '11 at 18:20
I wonder whats the use of `int (*a)[5]` if its not workable. — user743211, May 12 '11 at 18:24
It is just history from the C language. You can use arrays as parameter types, but it decays to pointers anyway. If you use C++ you should use a std::vector which knows its own size. — Bo Persson, May 12 '11 at 18:27
You may want to read our [array FAQ](http://stackoverflow.com/questions/4810664/). — fredoverflow, May 12 '11 at 18:58

score 3 · Answer 1 · answered May 12 '11 at 18:18

3

Using a function template would be better if you major emphasis is on passing the size of the array.

template <typename T, size_t N>
void Func(T (&a)[N])
{  
    for (int i = 0; i < N; i++)
      cout << a[i]; 
}

answered May 12 '11 at 18:18

Prasoon Saurav

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score 3 · Accepted Answer · answered May 12 '11 at 18:55

Firstly, the way you access the array is wrong. Inside your function you have a pointer to an array. The pointer has to be dereferenced first, before you use the [] operator on it

void Func(int (*a)[5])
{
   for (int i = 0; i < 5; i++)
      cout << (*a)[i];
}

Secondly, i don't understand why you need to "get" the size of the array, when the size is fixed at compile time, i.e. it is always 5. But in any case you can use the well-known sizeof trick

void Func(int (*a)[5])
{
   for (int i = 0; i < sizeof *a / sizeof **a; i++)
      cout << (*a)[i];
}

The sizeof *a / sizeof **a expression is guaranteed to evaluate to 5 in this case.

score 1 · Answer 3 · answered May 12 '11 at 18:45

It doesn't work because you didn't de-reference it enough. You have a pointer to an array- the pointer requires de-referencing, and the array needs indexing.

void Func(int (*a)[5])
{
   for (int i = 0; i < 5; i++)
      std::cout << (*a)[i];
}

int main()
{
   int a[] = {1, 2, 3, 4, 5};
   Func(&a);
}

score 0 · Answer 4 · answered May 12 '11 at 18:18

0

When you pass an array into a function, it degrades to a pointer and all size information is lost.

answered May 12 '11 at 18:18

jonsca

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score 0 · Answer 5 · answered May 12 '11 at 18:41

You can instead use a STL container, like vector, to make things easier.

#include <iostream>
#include <vector>
using namespace std;

void Func(vector <int> a){
   for (int i = 0; i < a.size(); i++)
   cout << a[i];
}

int main (){
       int nums[] = {1, 2, 3, 4, 5};
       vector<int> a (nums, nums + sizeof(nums) / sizeof(int) );
       Func(a);
}

score 0 · Answer 6 · answered May 12 '11 at 18:46

0

How do i get the size of array

You already know the size at compile-time, it's 5. You just forgot to dereference the pointer:

cout << (*a)[i];

answered May 12 '11 at 18:46

fredoverflow

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What about `int size = sizeof(*a)/sizeof(int);`? – user743211 May 12 '11 at 18:50
@user: You already told the compiler that you point to an array of 5 integers. What else do you hope to find out about the array? Yes, `sizeof(*a)/sizeof(int)` will yield 5, but again, you already know that. – fredoverflow May 12 '11 at 18:57

score -1 · Answer 7 · answered May 12 '11 at 18:18

-1

void Func(int * a)
{
   for (int i = 0; i < 5; i++)
      cout << *(a+i);
}

int main()
{
   int a[] = {1, 2, 3, 4, 5};
   Func(a);
}

answered May 12 '11 at 18:18

J T

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Passing an array by pointer

7 Answers7