default-arguments

问题 So the excellent answer to this question states that you can default template types in a forward declaration, however: You can specify each default template argument only once This is more fully specified in the linked question, but Boost's Property Tree stuff works with the ptree type: typedef basic_ptree<std::string, std::string> ptree; But the basic_ptree class is defined like this: template<class Key, class Data, class KeyCompare> class basic_ptree The only reason the ptree typedef is

问题 I expect the following code to compile: #include <iostream> template <class Tag = void, class T = int, class... Args> void print(T val = T{}, Args... args) { std::cout << val << ' ' << sizeof...(args) << std::endl; } int main() { print(); print(3.14); print(0, 1, 2); } While it compiles on GCC 5.2 (C++11) despite the unused-but-set-parameter warnings, clang 3.6 (C++11) gives the following error messages: main.cpp:4:33: error: missing default argument on parameter 'args' void print(T val = T{}

问题 I've just started going through a beginners book of C++. I have some java experience (but having said that, I've never used default arguments in java to be honest) So, as mentioned, my issue is with default arguments.. This is the code snippet I'm using: #include <iostream> using namespace std; //add declaration int add(int a, int b); int main (void) { int number1; cout << "Enter the first value to be summed: "; cin >> number1; cout << "\nThe sum is: " << add(number1) << endl; } int add(int a

问题 I've just started going through a beginners book of C++. I have some java experience (but having said that, I've never used default arguments in java to be honest) So, as mentioned, my issue is with default arguments.. This is the code snippet I'm using: #include <iostream> using namespace std; //add declaration int add(int a, int b); int main (void) { int number1; cout << "Enter the first value to be summed: "; cin >> number1; cout << "\nThe sum is: " << add(number1) << endl; } int add(int a

问题 This link doesn't answer my question so I'll ask it here: Basically I want to write a template function template <typename Out, typename In> Out f(In x); Here I always need to specify Out when calling f . I don't want to do it every time, so I basically want template <typename Out = In, typename In> Out f(In x); Which means if I don't specify Out , it will default to In . However, this is not possible in C++11. So my question is, is there any way to achieve the effect: calling f(t) will

问题 I ran into some trouble trying to implement a smart equality test macro-type template function in Visual C++ 2010 that had to do with a bug in VS in regard to default arguments of template functions. I fixed it by wrapping the value of the parameter in an extra function, but now I found that I can't use the function twice in one line! Header file: // example.h #pragma once #include <limits> namespace myspace { // Need to define this separately to avoid a Visual Studio bug template<typename T>

问题 public struct Test { public double Val; public Test(double val = double.NaN) { Val = val; } public bool IsValid { get { return !double.IsNaN(Val); } } } Test myTest = new Test(); bool valid = myTest.IsValid; The above gives valid==true because the constructor with default arg is NOT called and the object is created with the standard default val = 0.0. If the struct is a class the behaviour is valid==false which is what I would expect. I find this difference in behaviour and particularly the

问题 Suppose I have this code. Your basic "if the caller doesn't provide a value, calculate value" scenario. void fun(const char* ptr = NULL) { if (ptr==NULL) { // calculate what ptr value should be } // now handle ptr normally } and call this with either fun(); // don't know the value yet, let fun work it out or fun(something); // use this value However, as it turns out, ptr can have all kinds of values, including NULL, so I can't use NULL as a signal that the caller doesn't provide ptr. So I'm

问题 I have a C++ function that has 5 arguments, all of which have default values. If I pass in the first three arguments, the program will assign a default value to the last two arguments. Is there any way to pass 3 arguments, and skip one in the middle, giving values to say, the first, second, and fifth arguments? 回答1: Not directly, but you might be able to do something with std::bind: int func(int arg1 = 0, int arg2 = 0, int arg3 = 0); // elsewhere... using std::bind; using std::placeholders::

问题 I am trying to create this function below that takes one optional argument. It's an object, and the only prop needed is value however, I am looking to pass along any ...rest parameters that area also passed in. const useHook = <T extends { value: string }>({ value: v = '', ...props }: T = { value: '' } as T) => { const defaultValue = React.useMemo(() => v, [props]) const [value, setValue, resetValue] = Hooks.useState(defaultValue) const onChange = React.useCallback((e) => setValue(e.target