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Is it common to declare const pointers in C++?


Is it common to declare const pointers in C++?

By : Ryan Thies
Date : October 24 2020, 06:10 AM
Hope that helps A pointer is an object.
If you wish to prevent modification of the value held by the pointer, i.e. to make it hold only a single address within its lifetime, then yes make it const. The same considerations that apply for any other object type also apply here.
code :


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What use are const pointers (as opposed to pointers to const objects)?

What use are const pointers (as opposed to pointers to const objects)?


By : Ann
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further When you're designing C programs for embedded systems, or special purpose programs that need to refer to the same memory (multi-processor applications sharing memory) then you need constant pointers.
For instance, I have a 32 bit MIPs processor that has a little LCD attached to it. I have to write my LCD data to a specific port in memory, which then gets sent to the LCD controller.
How do you declare a const array of function pointers?

How do you declare a const array of function pointers?


By : David
Date : March 29 2020, 07:55 AM
around this issue Firstly, I've got functions like this. ,
Then I create my typedef for the array: void (*FP)();
code :
 void func1(){}
 void func2(){}
 void func3(){}

 typedef void (*FP)();


 int main()
 {
     const FP ar[3]= {&func1, &func2, &func3};
 }
 class x;
 typedef void (x::*FP)(); // you made a mistake here

 class x
 {
   public:
      void func1();
      void func2();
      void func3();
      static const FP array[3];
 };
Return a const vector of const shared pointers to const objects

Return a const vector of const shared pointers to const objects


By : Shashank Puthran
Date : March 29 2020, 07:55 AM
wish helps you I could be wrong but I really don't think you can achieve this.
shared_ptr can become a shared_ptr only through a construction of a new instance to shared_ptr
How to declare a member functions as const or not depending on whether a template parameter is const or not

How to declare a member functions as const or not depending on whether a template parameter is const or not


By : user2893573
Date : March 29 2020, 07:55 AM
hop of those help? , Eventually worked out how to do this using std::enable_if.
code :
template <typename T>
struct A {
  A(T a) : _value{a} {}
  template <typename Tp = T>
  std::enable_if_t<std::is_const<Tp>::value, T&> value() const {return _value;}
  template <typename Tp = T>
  std::enable_if_t<!std::is_const<Tp>::value, T&> value() {return _value;}
private:
  T _value;
};
template <typename T>
struct A {
  A(T a) : _value{a} {}
  template <typename Tp = T>
  typename std::enable_if<std::is_const<Tp>::value, T&>::type value() const {return _value;}
  template <typename Tp = T>
  typename std::enable_if<!std::is_const<Tp>::value, T&>::type value() {return _value;}
private:
  T _value;
};
Why I can not declare a variable inside an ES6 class by using let, const or var keyword but I can declare it directly?

Why I can not declare a variable inside an ES6 class by using let, const or var keyword but I can declare it directly?


By : MAH
Date : September 03 2020, 05:00 AM
this will help For the following code I wanted to know the reason behind this behavior in an ES6 Class:
code :
class One {
    //why the following code is not allowed.
    let check = false; 
    const PI = 3.14;   
    var v = 'Hello';    

    //why the following code is allowed.
    chk = false;       
    Pi = 3.14;         
    vv = "Hi";         
}
class One {
  chk = false;       
  Pi = 3.14;         
  vv = "Hi";         
}
class One {
  constructor() {
    this.chk = false;       
    this.Pi = 3.14;         
    this.vv = "Hi";         
  }
}
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