std::get_if
From cppreference.com
                    
                                        
                    
                    
                                                            
                    |   Defined in header  <variant>
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| (1) | (since C++17) | |
|   template <std::size_t I, class... Types> constexpr std::add_pointer_t<  | 
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|   template <std::size_t I, class... Types> constexpr std::add_pointer_t<  | 
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| (2) | (since C++17) | |
|   template <class T, class... Types> constexpr std::add_pointer_t<T> get_if(std::variant<Types...>* pv) noexcept;  | 
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|   template <class T, class... Types> constexpr std::add_pointer_t<const T> get_if(const std::variant<Types...>* pv) noexcept;  | 
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1) Index-based non-throwing accessor: If 
pv is not a null pointer and pv->index() == I, returns a pointer to the value stored in the variant pointed to by pv. Otherwise, returns a null pointer value. The call is ill-formed if I is not a valid index in the variant.2) Type-based non-throwing accessor: Equivalent to (1) with 
I being the zero-based index of T in Types.... The call is ill-formed if T is not a unique element of Types....Parameters
| I | - | index to look up | 
| Type | - | unique type to look up | 
| pv | - | pointer to a variant | 
Return value
Pointer to the value stored in the pointed-to variant or null pointer on error.
Example
Run this code
#include <variant> #include <iostream> int main() { std::variant<int, float> v{12}; if(auto pval = std::get_if<int>(&v)) std::cout << "variant value: " << *pval << '\n'; else std::cout << "failed to get value!" << '\n'; }
Output:
variant value: 12
See also
|    (C++17)  | 
  reads the value of the variant given the index or the type (if the type is unique), throws on error   (function template)  |