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{{Short description|C++ programming technique}}
'''Substitution failure is not an error''' ('''SFINAE''') refers to a situation in [[C++]] where an invalid substitution of [[template (programming)|template]] parameters is not in itself an error. David Vandevoorde first introduced the acronym SFINAE to describe related programming techniques.<ref>{{cite book | last=Vandevoorde | first=David |author2=Nicolai M. Josuttis | title=C++ Templates: The Complete Guide | publisher=Addison-Wesley Professional | year=2002 | isbn=0-201-73484-2}}</ref>▼
▲'''Substitution failure is not an error''' ('''SFINAE''')
Specifically, when creating a candidate set for [[overload resolution]], some (or all) candidates of that set may be the result of instantiated templates with (potentially deduced) template arguments substituted for the corresponding template parameters. If an error occurs during the substitution of a set of arguments for any given template, the compiler removes the potential overload from the candidate set instead of stopping with a compilation error, provided
==Example==
The following example illustrates a basic instance of SFINAE:
<
struct Test {
};
template <typename T>
void f(typename T::foo) {} // Definition #1
template <typename T>
void f(T) {}
int main() {
// thanks to SFINAE. return 0;
}
</syntaxhighlight>
Here, attempting to use a non-class type in a qualified name (<code>T::foo</code>) results in a deduction failure for <code>f<int></code> because <code>int</code> has no nested type named <code>foo</code>, but the program is well-formed because a valid function remains in the set of candidate functions.
Although SFINAE was initially introduced to avoid creating ill-formed programs when unrelated template declarations were visible (e.g., through the inclusion of a header file), many developers later found the behavior useful for compile-time introspection. Specifically, it allows a template to determine certain properties of its template arguments at instantiation time.
For example, SFINAE can be used to determine if a type contains a certain typedef:
<
#include <iostream>
template <typename T>
struct
// foobar. };
struct
};
int main() {
return 0;
}
</syntaxhighlight>
When <code>T</code> has the nested type <code>foobar</code> defined, the instantiation of the first <code>test</code> works and
== C++11 simplification ==
In [[C++11]], the above code could be simplified to:
<syntaxhighlight lang="cpp">
#include <iostream>
#include <type_traits>
template <typename T, typename = void>
struct HasTypedefFoobar : std::false_type {};
template <typename T>
struct HasTypedefFoobar<T, std::void_t<typename T::foobar>> : std::true_type {};
struct Foo {
using foobar = float;
};
int main() {
std::cout << std::boolalpha;
std::cout << HasTypedefFoobar<int>::value << std::endl;
std::cout << HasTypedefFoobar<Foo>::value << std::endl;
return 0;
}
</syntaxhighlight>
With the standardisation of the detection idiom in the [http://en.cppreference.com/w/cpp/experimental/lib_extensions_2 Library fundamental v2 (n4562)] proposal, the above code could be re-written as follows:
<syntaxhighlight lang="cpp">
#include <iostream>
#include <type_traits>
template <typename T>
using HasTypedefFoobarUnderlying = typename T::foobar;
struct Foo {
using foobar = float;
};
int main() {
std::cout << std::boolalpha;
std::cout << std::is_detected<HasTypedefFoobarUnderlying, int>::value << std::endl;
std::cout << std::is_detected<HasTypedefFoobarUnderlying, Foo>::value << std::endl;
return 0;
}
</syntaxhighlight>
The developers of [[Boost C++ Libraries|Boost]] used SFINAE in boost::enable_if<ref name="enable_if">[http://www.boost.org/doc/libs/release/libs/utility/enable_if.html Boost Enable If]</ref> and in other ways.
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{{reflist}}
{{C++ programming language}}
▲{{use dmy dates|date=January 2012}}
[[Category:C++]]
[[Category:Articles with example C++ code]]
[[Category:Software design patterns]]
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