Now, let's see an example using both auto and decltype: #include <iostream> #include <vector> int main(){ std::vector<int> vec(10); // using auto for type deduction for(auto i = vec.begin(); i < vec.end(); i++){ std::cin >> *i; } // using decltype for type deduction...
A Magic Spell: Using auto and decltype in C++11 to Declare Objects Without Spelling Out Their TypesDanny Kalev
decltype(ci) x = 0;// x has type const int decltype(cj) y = x;// y has type const int & and is bound to x decltype(cj) z;// error: z is a reference and must be initialized. 当表达式不是变量,而是可作为左值的对象时,那么decltype返回的时指向该对象类型的应用。 int*p = &i; d...
inta=3,b=4;decltype(a)c=a;// <=> int c = a; c 是int 型decltype(a=b)d=a;// <=>...
decltype(r+0) b;//ok: addition yields anint; b is an (uninitialized)intdecltype(*p) c;//error: c isint&andmust be initialized Here r is a reference, so decltype(r) is a reference type. If we want the type to which r refers, we can use r in an expression, such as r+0, ...
This is way better for the caller, and the template code loses nothing in readability--if anything, it's easier to understand! The joy of decltype and the new return value syntax Now you might be saying here--okay, that's great, but what if I wanted to *return* the value that this...
Shouldn't auto and decltype give the same results for the same input? They give different opinions on the type of an expression when this expression results in a reference value. Surely, the type of "" is the same, regardless of if I deduced it using au...
Use auto and decltype to declare a template function whose return type depends on the types of its template arguments. Or, use auto and decltype to declare a template function that wraps a call to another function, and then returns whatever is the return type of that other function. For ...
Yeah that's the thing. It's actually decltype(co_await this_coro::executor). It depends on the awaitable/promise. And that's encoded in the coro return type (thank C++ standards). You really do not want to know the type here. It's bad enough that the default be any_io_executor....
auto add(T t, U u) const -> decltype(t + u) {return t + u;}Known issues / limitationsWhen the compiler deduces the variable type, it uses the same rule as template argument deduction. However, it doesn’t support full template argument deduction rule. See the following example:...