In C , the "if" statement is commonly employed for conditional branching based on the outcome of a certain condition. However, for compile-time conditional evaluation, the "if constexpr()" statement serves as a potent tool. This article explores the distinctions between "if constexpr()" and "if()", highlighting their use cases and applicability.
The fundamental distinction between "if constexpr()" and "if()" lies in their timing of evaluation. "if constexpr()" evaluates the expression within its parentheses at compile-time, whereas "if()" evaluates its condition at runtime. This difference carries significant implications:
Consider a scenario where you have a function, "length", that requires different logic for determining the length of numbers and objects with a ".length()" function. Using "ifconstexpr()", you can handle both cases with a single function, as it allows you to evaluate the type of the value at compile-time. Here's an example:
template<typename T> auto length(const T& value) noexcept { if constexpr (std::is_integral<T>::value) { // is number return value; else return value.length(); }
Another use case of "if constexpr()" is to bypass potential errors arising from illegal operations. For instance, if you decide to invoke a member function on a value that may not have it, you can use "if constexpr()" to test the type and execute a different path if the function is unavailable:
template<typename T> bool contains(const T& value, const std::string& key) noexcept { if constexpr (has_find<T>::value) { // has std::find function return std::find(value.begin, value.end, key) != value.end(); else return false; }
Generally, "if constexpr()" should be employed when you need to make decisions based on the outcome of a condition during compilation. This allows the compiler to optimize the code and eliminate unnecessary or unreachable code. "if()", on the other hand, is appropriate when the condition's outcome is known only at runtime.
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