Programming Forums > Application Development > C++

Jessehk · Jul 16th, 2006, 10:16 PM

I recently came across this snippet of code:

 cpp Syntax (Toggle Plain Text) 
#include <iostream>
#include <cstddef>
#include <algorithm>
#include <iterator>
 
template <typename T, size_t size>
void print_all(const T (&arr)[size]) {
    std::copy(arr, arr + size, std::ostream_iterator<T>(std::cout, "\n"));
}
 
int main() {
    int arr[] = { 1, 2, 3, 4 };
 
    print_all(arr);
}
#include <iostream>
#include <cstddef>
#include <algorithm>
#include <iterator>

template <typename T, size_t size>
void print_all(const T (&arr)[size]) {
    std::copy(arr, arr + size, std::ostream_iterator<T>(std::cout, "\n"));
}

int main() {
    int arr[] = { 1, 2, 3, 4 };

    print_all(arr);
}

I'm particularly curious as to why this section:

 cpp Syntax (Toggle Plain Text) 
const T (&arr)[size]
const T (&arr)[size]

works. I have done some searching, but to no avail. Any info, or links to info documenting this feature of C++ would be appreciated. Thanks

Eoin · Jul 17th, 2006, 4:21 AM

This is a template in order to allow the function to work for all array types (well excluding dynamically created ones). If the function wasn't a template you would have to code one specifically for each array you'd use with it, for example;

 cpp Syntax (Toggle Plain Text) 
void print_all(const int (&arr)[4]) {
    std::copy(arr, arr + 4, std::ostream_iterator<int>(std::cout, "\n"));
}
void print_all(const int (&arr)[4]) {
    std::copy(arr, arr + 4, std::ostream_iterator<int>(std::cout, "\n"));
}

The

 cpp Syntax (Toggle Plain Text) 
const int (&arr)[4]
const int (&arr)[4]

is just the way you write the type for a const reference to an int array containing 4 elements.

Cache · Jul 17th, 2006, 12:25 PM

I think the question is more concerned with the specifics of how the second template parameter is deduced.

Cache · Jul 17th, 2006, 11:54 PM

Quote:

Originally Posted by Cache

I think the question is more concerned with the specifics of how the second template parameter is deduced.

Sorry to DP, but this has been bugging me, lol. The above quote should have read: I think the question is more concerned with the specifics of when (compile/link time vs runtime) the second template parameter is deduced.

grumpy · Jul 18th, 2006, 5:18 AM

All template parameters are "deduced" at compile time.

Jul 16th, 2006, 10:16 PM	#1
Jessehk The Oblivious One Join Date: May 2005 Location: Ontario, Canada Posts: 648 Rep Power: 4	Templates and Arrays I recently came across this snippet of code: cpp Syntax (Toggle Plain Text) #include <iostream> #include <cstddef> #include <algorithm> #include <iterator> template <typename T, size_t size> void print_all(const T (&arr)[size]) { std::copy(arr, arr + size, std::ostream_iterator<T>(std::cout, "\n")); } int main() { int arr[] = { 1, 2, 3, 4 }; print_all(arr); } #include <iostream> #include <cstddef> #include <algorithm> #include <iterator> template <typename T, size_t size> void print_all(const T (&arr)[size]) { std::copy(arr, arr + size, std::ostream_iterator<T>(std::cout, "\n")); } int main() { int arr[] = { 1, 2, 3, 4 }; print_all(arr); } I'm particularly curious as to why this section: cpp Syntax (Toggle Plain Text) const T (&arr)[size] const T (&arr)[size] works. I have done some searching, but to no avail. Any info, or links to info documenting this feature of C++ would be appreciated. Thanks __________________ Dr. Zoidberg: [ecstatic] I'm going to a movie... with FRIENDS!

Jul 17th, 2006, 4:21 AM	#2
Eoin Hobbyist Programmer Join Date: Jun 2006 Location: Ireland Posts: 152 Rep Power: 3	This is a template in order to allow the function to work for all array types (well excluding dynamically created ones). If the function wasn't a template you would have to code one specifically for each array you'd use with it, for example; cpp Syntax (Toggle Plain Text) void print_all(const int (&arr)[4]) { std::copy(arr, arr + 4, std::ostream_iterator<int>(std::cout, "\n")); } void print_all(const int (&arr)[4]) { std::copy(arr, arr + 4, std::ostream_iterator<int>(std::cout, "\n")); } The cpp Syntax (Toggle Plain Text) const int (&arr)[4] const int (&arr)[4] is just the way you write the type for a const reference to an int array containing 4 elements. __________________ Visit my website BinaryNotions.

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Jul 17th, 2006, 12:25 PM	#3
Cache Hobbyist Join Date: Sep 2005 Posts: 266 Rep Power: 4	I think the question is more concerned with the specifics of how the second template parameter is deduced.

Jul 18th, 2006, 5:18 AM	#5
grumpy Programming Guru Join Date: Jun 2005 Location: Adelaide, South Australia Posts: 1,260 Rep Power: 5	All template parameters are "deduced" at compile time.

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Programming Forums > Application Development > C++

Templates and Arrays