default-parameters

问题 It tried something like this, which doesn't work. Is there a way to get a similar effect? class A { public: int foo(); void bar(int b = foo()); }; 回答1: Yes. Overload the function and call the member-function in it. void bar() { bar(foo()); } 来源： https://stackoverflow.com/questions/4901233/is-there-anyway-to-use-a-member-function-as-a-default-parameter

问题 Is there any way to specify a default parameter in a variadic function?(Applies to templates also) 回答1: In C++ you can replace the variadic function with one based on the Named Parameter Idiom. See the C++ FAQ item 10.20 What is the "Named Parameter Idiom"?. That gives you default functionality & convenient notation. Cheers & hth., 回答2: Why would you need both variadic and default params? For example, myFunc(int a=5, int b=5, int c=5); can receive 0-3 parameters with default values, and

问题 The following line has the error Default argument is not allowed . public ref class SPlayerObj{ private: void k(int s = 0){ //ERROR } } Why C++ has no default argument on managed types ? I would like to know if there is a way to fix this. 回答1: It does have optional arguments, they just don't look the same way as the C++ syntax. Optional arguments are a language interop problem. It must be implemented by the language that makes the call, it generates the code that actually uses the default

问题 I have the following class: template <typename Type = void> class AlignedMemory { public: AlignedMemory(size_t alignment, size_t size) : memptr_(0) { int iret(posix_memalign((void **)&memptr_, alignment, size)); if (iret) throw system_error("posix_memalign"); } virtual ~AlignedMemory() { free(memptr_); } operator Type *() const { return memptr_; } Type *operator->() const { return memptr_; } //operator Type &() { return *memptr_; } //Type &operator[](size_t index) const; private: Type *memptr

问题 My Javascript function leads my console to return me : TypeError: style is null Here the snippet: let style = { one: 1, two: 2, three: 3 } function styling(style = style, ...ruleSetStock) { return ruleSetStock.map(ruleSet => { console.log(ruleSet) return style[ruleSet] }) } console.log(styling(null, "one", "two", "three")) I can't understand why. It seems to me everything is great, Any hint would be great, thanks. 回答1: Default parameters is assigned only if no value or undefined is passed let

问题 Today I was creating a default parameter value in a constructor. public SomeClass (String something = String.Empty) { // ... } The compiler complained. Default parameter value for "something" must be a compile-time constant. I was under the impression that Empty on the String class was a compile-time constant. .field public static initonly string Empty Am I missunderstanding the meaning of compile-time constant, or is it just more wackyness that I need to accept? 回答1: The accepted answer to

问题 This question already has answers here : Nil cannot be assigned to type ()->()? (3 answers) Swift 3 optional escaping closure parameter (5 answers) Closed 2 years ago . Thanks to the marvels of Swift, we can have: func someFunction(someParameter: someType, someOtherParameter: someOtherType) We call it like so: someFunction(x, y) We also have this: func someFunction(someParameter: someType, someOtherParameter: someOtherType?) We may call this like so: someFunc(x, y) OR someFunc(x, nil) However

问题 When I try to compile and run this code (only the first three lines really matter): class object; object getObject(); void doSomething(object o = getObject()); class object{ public: int num = 0; }; object getObject(){ return {}; } void doSomething(object o){ o.num = 5; } int main(){} I get this error: main.cpp:3:39: error: invalid use of incomplete type 'class object' void doSomething(object o = getObject()); ^ main.cpp:1:7: note: forward declaration of 'class object' class object; ^ How

问题 I have trouble understanding how default parameters interact with multiple clauses in named functions. It boils down to, why does the following snippet work? defmodule Lists do def sum([], total \\ 0), do: total def sum([h|t], total), do: h + sum(t, total) end From my understanding this gets expanded by the compiler into: defmodule Lists do def sum([]), do: sum([], 0) def sum([], total), do: total def sum([h|t], total), do: h + sum(t, total) end So I would expect the following to happen: iex

问题 Following code can't be compiled with g++ version 5.4.0 with option -std=c++1y : void f(int=0) ; int main() { f(); // ok (*f)(2);// ok (*f)();// ok c++11; error with c++14: too few arguments to function return 0; } The function declared to have default argument, so what is wrong here? thanks for help. And why does g++ -c -std=c++11 compile? 回答1: Accepting (*f)() as valid is a GCC bug. The letter of the standard indicates that using a function name with unary * should cause the function name