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:: scope resolution operator in front of a template function call in c++

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问题 I'm stuck with templates and scope resolution operator. I found these line in a file, I'm not able to figure out why we are using :: in front of a template function call, as of my knowledge we can only use :: in front of variables when refering to a global variable. Any Idea will be helpful #define CREATE_AND_DECODE_TYPE(Type, buffer, pType) \ ::CreateAndDecodeType<Type>(buffer, pType, throwVarBindExceptions, static_cast<Type *>(NULL)) 回答1: The scope resolution operator :: (at the beginning)

:: scope resolution operator in front of a template function call in c++

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Scope-resolution operator :: versus member-access operator . in C#

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问题 In C#, what's the difference between A::B and A.B ? The only difference I've noticed is that only :: can be used with global , but other than that, what's the difference? Why do they both exist? 回答1: with :: you can do things like... extern alias X; extern alias Y; class Test { X::N.A a; X::N.B b1; Y::N.B b2; Y::N.C c; } and there are times when . is ambiguous so :: is needed. here's the example from the C# language spec namespace N { public class A {} public class B {} } namespace N { using

whats the difference between dot operator and scope resolution operator

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问题 I just wanted to know the difference between . operator and :: operator? 回答1: The former (dot, . ) is used to access members of an object, the latter (double colon, :: ) is used to access members of a namespace or a class. Consider the following setup. namespace ns { struct type { int var; }; } In this case, to refer to the structure, which is a member of a namespace, you use :: . To access the variable in an object of type type , you use . . ns::type obj; obj.var = 1; 回答2: Another way to

Is there a scope resolution operator in C language?

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问题 I am reading a book on the C language ( 'Mastering C' ), and found the topic on scope resolution operator ( :: ) on page 203, on Google Books here. But when I run the following code sample (copied from the book), the C compiler gives me an error. I searched on the internet but I am unable to find any reference to a scope resolution operator in C. #include <stdio.h> int a = 50; int main(void) { int a =10; printf("%d",a); printf("%d\n", ::a); return 0; } So if I want to access a global variable

struct with member function as parameter

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问题 I am a beginner in C++ and stack exchange. I am working on an Interface class that gets keyboard input and checks to see whether it is correct through looping through an array of structs which contains strings to compare to and strings to output depending if it is equal to the compare string or not. If the input is correct, it will print the string within the struct and a function within the structure is called and does some action. interface.hpp #include <string> class Input_Interface {

Call private method from inherited class

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问题 I want to implement a hook-system in my simple ORM, in PHP: class Record { public function save() { if (method_exists($this,"before_save")) { $this->before_save(); } //...Storing record etc. } } class Payment extends Record { private function before_save() { $this->payed_at = time(); } } $payment = new Payment(); $payment->save(); This results in a Fatal Error: Fatal error: Call to private method Payment::before_save() from context 'Record' in Makes sense. I could change the scope to public,

Call private method from inherited class

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I want to implement a hook-system in my simple ORM, in PHP: class Record { public function save() { if (method_exists($this,"before_save")) { $this->before_save(); } //...Storing record etc. } } class Payment extends Record { private function before_save() { $this->payed_at = time(); } } $payment = new Payment(); $payment->save(); This results in a Fatal Error: Fatal error: Call to private method Payment::before_save() from context 'Record' in Makes sense. I could change the scope to public, but that seems ugly: no-one but Payment has anything to do with before_save() . It is best left private

vim does not find and replace simple phrase that is clearly present

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问题 I have a simple vim problem that Google hasn't managed to help me with. Any thoughts are appreciated. I do the following search and replace: :s/numnodes/numnodes1/g On a file containing the following text: numprocs=0 numnodes=0 I get E486: Pattern not found The position of the green square which indicates where I'd start typing is clearly above the pattern. I tried searching for other short phrases not involving regex, which are also present, which also fail. A simple /numnodes highlights

What does C++ syntax “A::B:A {};” mean

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问题 What does C++ syntax struct A::B:A {}; mean? Where is this name definition (or access) described in the C++ standard? #include <iostream> struct B; struct A { struct B; }; struct A::B:A { }; int main() { A::B::A::B b; std::cout<<"Sizeof A::B::A::B is " << sizeof(A::B::A::B)<<std::endl; return 0; } 回答1: This definition struct A { struct B; }; Defines a struct A with a declaration of a nested struct B 1 . The fully qualified name of B is A::B , you could say B is inside the "namespace" of A .