C++ Constructor Initialization

Question

I have the following code:

    struct Y {
        string& s1; //string s1; throws no error
        Y( string src ) : s1(src) { cout<<"s1: "<<s1<<endl; }
        void show(){ cout<<s1<<endl; }
    };

    int main()
    {
        Y y1("Krypton");
        y1.show(); //run-time error
    }

y1.show() should display "Krypton", but I get a runtime error (due to s1 being uninitialized when y1.show() is being called?).

Q1. Why would s1 be uninitialized when it has already been initialized in the constructor initialization list? Q2. Why don't I get the same error if I use string s1; instead of the reference?

Any help would be greatly appreciated.

Regards,

Jay.

It is usually a good idea to describe the actual error, rather than just saying "I get a runtime error". — Stack Overflow is garbage
– Stack Overflow is garbage, Commented Jul 7, 2013 at 9:47

David Schwartz · Accepted Answer · 2013-07-07 09:39:45Z

1

You are initializing a reference from a temporary. If you want instances of struct Y to hold an actual string, rather than just a reference, you need a member variable of type string, not reference to string.

Q1. Why would s1 be uninitialized when it has already been initialized in the constructor initialization list?

It was initialized, but to be a reference to a string that no longer exists.

Q2. Why don't I get the same error if I use string s1; instead of the reference?

If the class contains a string, then it doesn't contain a reference to a string that exists.

What string do you think the reference is a reference to?

answered Jul 7, 2013 at 9:39

David Schwartz

184k18 gold badges229 silver badges293 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

VivereJay Over a year ago

Thanks. So if I understand this correctly, the reference member pointed to a value that did not exist once the constructor execution was complete. But reference variables have to be initialized in the initialization list. In that case, how can I initialize s1 correctly?

David Schwartz Over a year ago

It ceased to exist as soon as this line of code completed Y y1("Krypton");. Because you passed the constructor a const char *, it had to create a temporary string. That temporary ceased to exist as soon as it was no longer needed. When you take a reference to an object, it is your responsibility to ensure the reference isn't dereferenced after the object ceases to exist!

StoryTeller - Unslander Monica · Accepted Answer · 2013-07-07 09:39:42Z

1

s1 is set to reference src, which is a temporary object and is destroyed when the constructor exists.

When you make s1 a string, it is a local copy that doesn't depend on src

answered Jul 7, 2013 at 9:39

StoryTeller - Unslander Monica

172k22 gold badges401 silver badges481 bronze badges

Comments

superK · Accepted Answer · 2013-07-07 10:03:56Z

0

Because your constructor Y(string src) accepted a template value, and s1 reference to this template value, and then you call show() print the value witch is destroty after constructor.

Answer:

1.Because the value s1 referenced is destroy after constructor.

2.If you use string s1 instead of string& s1, witch assign src yo s1 when call constructor, and show() will print the member value of s1.

If you want to pring "Krypton", you can change Y( string src ) to Y( string& src )

edited Jul 7, 2013 at 10:03

answered Jul 7, 2013 at 9:57

superK

4,0028 gold badges32 silver badges56 bronze badges

Comments

john_zac · Accepted Answer · 2013-07-07 12:15:26Z

0

In your case you can initialize the temporary member variable like this

struct Y 
{
    string& s1;

    Y( string& src )
        : s1(src)
    {
        cout<<"s1: "<<s1<<endl;
    }

    void show()
    {
        cout<<s1<<endl;
    }
};

int main()
{
    string s1_real = "Krypton";
    Y y1(s1_real);
    y1.show();

    return 0;
}

Note that

Constructor takes a reference (string& src not string src). Because if you use "string src" it will be a temporary variable which will be destroyed as soon as the function stack unwinds and s1 ends up being a dangling reference. If you use "string& src" there is no new memory location created; src points to s1_real and therefore s1 ends up pointing to s1_real.
You create a variable (s1_real) which is used to create y1. This is because you can't take a reference of a temporary variable (you can only take a const reference, in which case src and s1 too must be declared as const references)

Always note that the lifetime of y1 must be less than that of s1_real or else y1.s1 ends up being a dangling reference

answered Jul 7, 2013 at 12:15

john_zac

1,2912 gold badges14 silver badges16 bronze badges

1 Comment

VivereJay Over a year ago

Thanks John! I'd had the same line of thought, however had missed changing the ctor to accept reference to the string defined in main(). So it was still crashing, again due to all of the reasons explained above :)

Collectives™ on Stack Overflow

C++ Constructor Initialization

4 Answers 4

2 Comments

Comments

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

2 Comments

Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related