Possible Duplicate:
Sizeof array passed as parameter
Given the function below I understand that sizeof returns the size of the pointer of the type in the array.
int myFunc(char my_array[5])
{
return sizeof(my_array);
}
However calling sizeof on an array not passed as a parameter normally returns the sizeof the array.
What causes this inconsistency? What is the best way of getting the size of an array when it is passed as a parameter?
You have to pass the array by reference; you cannot pass arrays by value (they decay into pointers to the first element):
int myFunc(char const (& my_array)[5])
{
return sizeof(my_array); // or just "return 5"
}
In case that you want to call this function with a reference to lots of different local arrays, you can use a function template instead to have the array size deduced:
template <unsigned int N>
int myFunc(char const (& my_array)[N])
{
return sizeof(N); // or "return N"
}
You should really make the functions return an unsigned int or an std::size_t, though. In C++11, you can also declare them as constexpr.
What causes this inconsistency?
The name of the array decays as an pointer to its first element.
When you pass an array to an function, this decaying takes place and hence the expression passed to sizeof is a pointer thus returning pointer size.
However, an array passed to sizeof always returns size of the array because there is no decaying to an pointer in this case.
What is the best way of getting the size of an array when it is passed as a parameter?
Don't pass them by value, pass them by reference or
Pass size as an separate argument to the function.
When you pass an array to a function you must call the function same as follow : myFun(my_array,size), where the size is an integer variable contains size of the array.(here it's myFun(my_array,5)) . and the signature of the function must be : int myFunc(char my_array[], int size)
