How does a member initialization list work with char pointers?

Question

I understand that a member initialzation list can be used to initialize objects and variables that have a known size.

However, how does a member initialization list actually work with pointers, in particular const char* pointers, as in MyClass below? Will it simply point to the memory location of the const char* which I pass in?

class MyClass
{
  public:
    MyClass(const char* str): _str(str)
    {
    }

  private:
    const char* _str;
};

*It will simply point to the memory location of the const char* which I pass?* Yep. What behavior are you trying to get? Do you want `Myclass` to have a copy of the passed in c-string? — NathanOliver, Jun 25 '21 at 19:41
It’s not clear what you’re referring to. The code you posted does not involve initializer lists. Did you mean class member initializers? — Sneftel, Jun 25 '21 at 19:42
@Engineer999 The both pointers str and _str will point to the same object. — Vlad from Moscow, Jun 25 '21 at 19:46
If you want the text copied: 1) Allocate same or larger space as text pointed to by `str`; 2) Copy the text using `strcpy` or `memcpy`. — Thomas Matthews, Jun 25 '21 at 20:03

score 1 · Answer 1 · answered Jun 25 '21 at 19:46

1

Yes, it will simply initialize _str with the value of str - ie, make _str point to the same memory address that str is pointing at.

answered Jun 25 '21 at 19:46

Remy Lebeau

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score 0 · Answer 2 · edited Jun 25 '21 at 19:47

I understand that a member initialzation list can be used to initialize objects and variables that have a known size.

Member initialization lists can be used to initialize members and base sub objects. Those always have a known size.

However, how does a member initialization list actually work with pointers

All pointers have a known size.

Will it simply point to the memory location of the const char* which I pass in?

Yes. That's what a pointer value is.

How does a member initialization list work with char pointers?

2 Answers2