Sized array in function signature [duplicate] - c++

Question

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What is the purpose of sized array as function argument is c and c++?
(2 answers)
Closed 1 year ago.
Does defining a sized array in a function signature (as opposed to the more commonly used unsized array or pointer syntax) have any bearing at all? My compiler is ignoring it completely, as the following sample code shows (which runs, although it prints some garbage values when a smaller-sized array is passed).
#include <iostream>
using namespace std;
void printArray(int intArray[5]) {
for (int i = 0; i < 5; i++) {
cout << intArray[i] << " ";
}
cout << endl;
}
int main()
{
int array1[1] = {1}; // Smaller array size than in the function signature
cout << "\nInvocation 1\n";
printArray(array1);
int array2[4] = {1, 2, 3, 4}; // Smaller array size than in the function signature
cout << "\nInvocation 2\n";
printArray(array2);
int array3[8] = {1, 2, 3, 4, 5, 6, 7, 8}; // Larger array size than in the function signature
cout << "\nInvocation 3\n";
printArray(array3);
return 0;
}

Answer 1

Does defining a sized array in a function signature (as opposed to the more commonly used unsized array or pointer syntax) have any bearing at all?
No, not the way you do it.
This would on the other hand make intArray a reference to the array you pass as an argument:
void printArray(int (&intArray)[5]) {
for (int i = 0; i < 5; i++) {
cout << intArray[i] << " ";
}
cout << endl;
}
A more generic form would be to make it a function template. This would then be instantiated once for each array size you call it with:
template<std::size_t N>
void printArray(int (&intArray)[N]) {
for (int i = 0; i < N; i++) {
cout << intArray[i] << " ";
}
cout << endl;
}

Answer 2

Does defining a sized array in a function signature (as opposed to the more commonly used unsized array or pointer syntax) have any bearing at all?
No, it has not, the passed array argument will always decay to a pointer to its first element, placing a size is indeed pointless from the compiler standpoint, it will ignore it, having void printArray(int intArray[5]){...}, void printArray(int intArray[]){...} or void printArray(int* intArray){...} will be basically the same.
My compiler is ignoring it completely, as the following sample code shows (which runs, although it prints some garbage values when a smaller-sized array is passed).
Which makes sense because in the loop you are accessing elements outside the bounds of the array, so the behavior is undefined.

Answer 3

This is how I would do it using std::array
#include <array>
#include <iostream>
template<typename type_t, std::size_t N>
void printArray(const std::array<type_t,N>& values)
{
for (const auto& value : values)
{
std::cout << value << " ";
}
std::cout << std::endl;
}
void printArray_s(const std::array<int, 5>& values)
{
for (const auto& value : values)
{
std::cout << value << " ";
}
std::cout << std::endl;
}
// helper function to avoid having to type value of N in std::array (syntactic sugar)
// used for array3
template<typename array_type, typename... args_t>
constexpr auto make_array(const args_t&&... values)
{
return std::array<array_type, sizeof...(args_t)>{ static_cast<array_type>(values)... };
}
int main()
{
// Smaller array size than in the function signature
std::array<int,1> array1{ 1 };
std::cout << "\nInvocation 1\n";
//printArray_s(array1); // <== will correctly give compiler error
// Smaller array size than in the function signature
std::array<int, 4> array2{ 1, 2, 3, 4 };
std::cout << "\nInvocation 2\n";
printArray(array2);
// Larger array size than in the function signature
auto array3 = make_array<int>(1, 2, 3, 4, 5, 6, 7, 8 );
std::cout << "\nInvocation 3\n";
printArray(array3);
return 0;
}