What is the difference between these C++ codes?

Question

I was trying this constructor code and got confused with how it works..

#include<iostream>
using namespace std;
class XYZ{
    int a,b;
    public:
        XYZ(int i, int j):a(i), b(a*j){}
        void show(){
            cout<<a<<"\t"<<b<<"\n";
        }
};
int main(){
    XYZ X(2,3);
    X.show();
    return 0;
}

it gives the expected results but,

#include<iostream>
using namespace std;
class XYZ{
    int b,a;
    public:
        XYZ(int i, int j):a(i), b(a*j){}
        void show(){
            cout<<a<<"\t"<<b<<"\n";
        }
};
int main(){
    XYZ X(2,3);
    X.show();
    return 0;
}

give unexpected result.

the only difference is int a,b and int b,a

and how XYZ(int i, int j):a(i), b(a*j){} works exactly?

One question per question please. I've told you that several times before. Also, after almost four years and 20 questions, could you consider getting into the habit of _properly formatting your posts_ please? Full stops, capital letters... Thanks. — Lightness Races in Orbit, Apr 09 '16 at 18:36
Possible duplicate of [What's the point of g++ -Wreorder?](http://stackoverflow.com/questions/1828037/whats-the-point-of-g-wreorder) — iksemyonov, Apr 09 '16 at 18:39
Use `-Wall` to catch this (`-Wreorder` is the required flag, to be more precise, but it's included in `-Wall`). — iksemyonov, Apr 09 '16 at 18:40

Benjamin Lindley · Answer 1 · 2016-04-09T18:31:56.653

6

Well, you already spotted the difference. And it's quite an important one. Members of a class are initialized in the order in which they appear in the class definition, regardless of the order they appear in the constructor's member initializer list. And since the initialization of b depends on the value of a, it is important that a should be initialized first. That's why this works:

int a,b;

But this doesn't:

int b,a;

It would be better though(much less error prone) if you write your constructor such that these dependencies don't exist. Then you don't have to worry about the order of declaration.

XYZ(int i, int j):a(i), b(i*j){}

edited Apr 09 '16 at 18:31

answered Apr 09 '16 at 18:27

Benjamin Lindley

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I m new to OOPs, Can u explain this line `XYZ(int i, int j):a(i), b(a*j){}` – Abhishek Sharma Apr 09 '16 at 18:30
actually there should be a compiler warning for this – 463035818_is_not_an_ai Apr 09 '16 at 18:33
2

@AbhishekSharma: http://stackoverflow.com/questions/1711990/what-is-this-weird-colon-member-syntax-in-the-constructor – Benjamin Lindley Apr 09 '16 at 18:35

score 5 · Answer 2 · edited Apr 09 '16 at 18:27

5

On your second example, b is initialized before a. So b gets the result of evaluating a*j, where a is still uninitialized.

edited Apr 09 '16 at 18:27

Barry

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answered Apr 09 '16 at 18:26

Markus

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What is the difference between these C++ codes?

2 Answers2

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