CODEWITHSUNDEEP
javaoperator-keywordoperator-precedence
staroski
staroski

Reputation: 3

Java postfix operator precedence in expressions with method calls

My doubt is why in the following code the output is 2 and 1 respectively? Is this really OK? For my perception, the method 'm' should receive the value 1 since it's used the postfix operator on variable 'i' instead of the prefix operator.

public class PostfixDoubt {

    public static void main(String[] args) {
        int i = 1;
        // why does m receive 2 as argument and not 1?
        i = i++ + m(i);
        System.out.println(i);
    }

    public static int m(int i) {
        System.out.println(i);
        return 0;
    }
}

Bellow is the bytecode decompiled with javap:

public class PostfixDoubt {
  public PostfixDoubt();
    Code:
       0: aload_0
       1: invokespecial #8                  // Method java/lang/Object."<init>":()V
       4: return

  public static void main(java.lang.String[]);
    Code:
       0: iconst_1
       1: istore_1
       2: iload_1
       3: iinc          1, 1
       6: iload_1
       7: invokestatic  #16                 // Method m:(I)I
      10: iadd
      11: istore_1
      12: getstatic     #20                 // Field java/lang/System.out:Ljava/io/PrintStream;
      15: iload_1
      16: invokevirtual #26                 // Method java/io/PrintStream.println:(I)V
      19: return

  public static int m(int);
    Code:
       0: getstatic     #20                 // Field java/lang/System.out:Ljava/io/PrintStream;
       3: iload_0
       4: invokevirtual #26                 // Method java/io/PrintStream.println:(I)V
       7: iconst_0
       8: ireturn
}

Upvotes: 0

Views: 245

Answers (1)

Sotirios Delimanolis
Sotirios Delimanolis

Reputation: 279990

This operation

i++

increments the value stored in the i variable and returns its previous value. When i is re-evaluated to be passed as an argument to 

m(i)

the new, incremented, value is used.

int i = 1;
i = i++ + m(i);

looks like

1: 1 (i = 2) + m(i)
2: 1 (i = 2) + m(2)
3: 1 (i = 2) + Whatever value m(2) returns
4: whatever value is the result of that addition
5: that value is assigned to i

Upvotes: 3

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