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C++模板

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C++模板推导规则

模板推导规则

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template<typename T>
void f(ParamType param);

f(expr); // deduce T and ParamType from expr

在编译期间，编译器通过expr推导T和ParamType的类型。直觉上，T的类型总是与expr一致，然而实际上，T的类型推到结果，不仅仅依赖于expr的类型，还取决于ParamType的形式。

ParamType 可以写为 T& / T&& / T* 分为为 引用/万能引用/指针

指针或引用

若expr是引用（左值引用或右值引用），则先将引用部分忽略；然后，对expr的类型和ParamType的类型进行模式匹配，来决定T的类型

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template<typename T>
void f(const T& param); // ParamType 为左值引用的形参

int x = 27;
int &rx = x;
const int &crx = x;
f(x); // T is int, param's type is const int&
f(rx); // T is int, param's type is const int&
f(crx); // T is int, param's type is const int&

const int y = 28;
f(y); // T is int, param's type is const int&

int &&rr = 27;
f(rr); // T is int, param's type is const int&

const int &&crr = 28;
f(crr); // T is int, param's type is const int&

int *p = &x;
const int *cp = &x;
const char * const ccp = "abcd";
f(p); // T is int*, param's type is int* const &
f(cp); // T is const int*, param's type is const int* const &
f(ccp); // T is const char*, param's type is const char* const &

f(27); // T is int, param's type is const int &