为什么首先调用 Overrided 函数? [英] Why the Overrided function getting called first?

问题描述

I executed the following program and I am curious about the output i got in which the function output is getting printed first even if it was the variable i tried to print first.

class Baap{

    public int h = 4;
    public int getH(){
        System.out.println("Baap "+h); return h;
    }

}

public class Beta extends Baap{

    public int h = 44;
    public int getH(){
        System.out.println("Beta "+h); return h;
    }
    public static void main(String args[]){
        Baap b = new Beta();
        System.out.println(b.h+" "+b.getH());
        Beta bb = (Beta)b;
        System.out.println(bb.h+" "+bb.getH());
    }

}

The output was as follows

Beta 44
4 44
Beta 44
44 44

Can somebody help me understand why the function block gets executed first?

解决方案

Your System.out.println line prints a String.

The String is evaluated at run-time as b.h + " " + b.getH(), so it concatenates b.h, space and the result of the method b.getH(), so it calls getH() which prints Beta 44, then prints the result 4 44.

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