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);
}
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.
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];
}
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);
}
When you pass an array into a function, it degrades to a pointer and all size information is lost.
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);
}
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];
int size = sizeof(*a)/sizeof(int);?
sizeof(*a)/sizeof(int) will yield 5, but again, you already know that.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);
}
int (*a)[5]if its not workable.