为什么在对同一个字符数组使用 sizeof 时会得到不同的结果？

Question

In main class I print sizeof(person->name) and then I do sizeof(name) which are same as I pass same char array to Person constructor. But why I get varied results in both cases
in first case sizeof returns 32 whereas in second case sizeof return 6 This is output

This is the code :-

#include <iostream>
#include "Person.h"

int main()
{

    char name[] = {'H','o','b','b','i','t'};
    Person *person = new Person(name , 203);

    std::cout << "p->Name size - " << sizeof(person->name) << " char array size   " << sizeof(name) << std::endl;

    delete person;

    return 0;

}

#include "Person.h"
#include <iostream>

Person::Person(){};

Person::Person(char name[],int age)
{
    this->name = name;
    this->age = age;
}

Person::~Person()
{
    std::cout << "\n Destructor called" << std::endl;
}

void Person::sayHello()
{
    std::cout << "\n Hello " << this->name << " " << this->age << std::endl ;   
}

Answer 1

sizeof in a std::string does not measure the complete amount of memory the string occupies. It measures how large the std::string object is.

The sizeof on a std::string is a constant value that is independent on the length of the string that is stored in it.

A std::string does have an overhead compared to a char name[] due to meta information and small string optimizations, and that overhead is (depending on the implementation) at a maximum of around 20 bytes.

For a range of 0 - ~20 chars a std::string will always occupy ~20 bytes memory. For more than ~20 chars the std::string will occupy ~20 + number of char bytes.

So yes for a really small number of chars it can be 5 times more bytes. But for a normal use case, that overhead can be ignored.