Java中println（）方法的奇怪行为

Question

class W {
    static int count=0;

    W() {
        count++;
        System.out.print("c ");
    }

    public static void main(String[] args) {
        System.out.println(new W().count+" "+new W().count);
    }
}

Expected output

c 1 c 2

Actual output

cc 1 2

Why?

Answer 1

In this line:

System.out.println(new W().count+" "+new W().count);

The instances of W are instantiated prior to the evaluation of the rest of the statement.

The order of operations is:

1. Instantiate first new W(), causing c to be printed
2. Evaluate first new W().count
3. Instantiate second new W(), causing c to be printed
4. Evaluate second new W().count
5. Concatenate the result and print.

Answer 2

The actual order of things executed by the JVM is as follows:

1. 1st W object is instantiated and its count property read.

   Here the first c is send to output.

2. 2nd W object is instantiated and its count property read.

   Here the second c is send to output.

3. The string argument for System.out.println() is built. ( == "1 2" )

4. The string argument is sent to the output.

Thus the output results to cc 1 2 .

Answer 3

This is because the System.out.println statement in the constructor is executed first and then only the calling System.out.println executed. Remember the System.out.pritnln in constructor executed completely before the calling System.out.println The order is

• new W() causes System.out.println() to be called hence printing "C"
• again new W() causes System.out.println() to be called hence printing "C"
• After this is complete, outer System.out.println prints new count.

Answer 4

main中的println方法需要知道它的参数是什么，所以JVM首先构造对象然后将它们发送到println方法

Answer 5

class W {
static int count=0;

W() {
    count++;
    System.out.print("c ");
}

public static void main(String[] args) {
    System.out.println(new W().count+" "+new W().count);
}

Because new W() will call the constructor W() then it will print "C" out, after that the count become 1 then you call new W() again, so another "C" is here. Finally, you call system.out.println(1 + " " + 2). then the 1 2 is in the output buffer.

Answer 6

class W {
    static int count=0;

    W() {
        count++;
        System.out.print("c ");
    }

    public static void main(String[] args) {
        //modify System.out.println(new W().count+" "+new W().count);
        System.out.print(new W().count + " ");
        System.out.print(new W().count);
    }
}

OUTPUT

c 1 c 2

The actual order of things executed by the JVM is as follows:

1st W object is instantiated and its count property read.

Here the first c is send to output.

2nd W object is instantiated and its count property read.

Here the second c is send to output.

The string argument for System.out.println() is built. ("1 2")

The string argument is sent to the output.

Thus the output results to cc 1 2.

Answer 7

class W {
    static int count=0;

    W() {
        count++;
        System.out.print("c ");
    }

    public static void main(String[] args) {
        System.out.println(new W().count+" "+new W().count);
    }
}

this is to do with the operator priority in java.

The new operator has higher precedence over the concatenation that's why the second new operator gets priority over the concatenation operator and the second c is getting print. after the second call to new W() the Statement will look like.

 System.out.println(1 +""+2);

and the output will be :- cc 1 2