std::get_if (std::variant)
Defined in header <variant>
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(1) | (since C++17) | |
template
<
std::size_t I, class... Types
>
constexpr
std::add_pointer_t
<
std::variant_alternative_t
<I, std::variant
<Types...>>>
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||
template
<
std::size_t I, class... Types
>
constexpr
std::add_pointer_t
<
const
std::variant_alternative_t
<I, std::variant
<Types...>>>
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(2) | (since C++17) | |
template
<
class T, class... Types
>
constexpr std::add_pointer_t<T> |
||
template
<
class T, class... Types
>
constexpr
std::add_pointer_t
<
const T>
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||
I
I
being the zero-based index of T
in Types.... The call is ill-formed if T
is not a unique element of Types...
Template parameters
I | - | index to look up |
Type | - | unique type to look up |
Parameters
pv | - | pointer to a variant |
Return value
Pointer to the value stored in the pointed-to variant or null pointer on error.
Example
#include <iostream> #include <variant> int main() { auto check_value = [](const std::variant<int, float>& v) { if (const int* pval = std::get_if<int>(&v)) std::cout << "variant value: " << *pval << '\n'; else std::cout << "failed to get value!" << '\n'; }; std::variant<int, float> v{12}, w{3.f}; check_value(v); check_value(w); }
Output:
variant value: 12 failed to get value!
See also
(C++17)
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reads the value of the variant given the index or the type (if the type is unique), throws on error (function template) |