Java - Why the following code print out "BAC", instead of "ABC"?

Question

Please help me understand this code. I am new to java.

// C.java
class C { 
  public static void main(String arg[]) { 
    System.out.println("A"+new C()); 
  } 
  public String toString() { 
    System.out.print("B"); 
    return "C"; 
  } 
}
// output:
// BAC  

Answer 1

The evaluation goes something like:

Call println("A" + new C())
Since new C() hasn't been computed yet, we need to compute it, so...
Compute new C().toString()
    Print "B"
Print line with "A" + "C"

As you can see, the order of the print statements is "B", "A", "C"

Answer 2

You need understand 2 concepts here: Java left-to-right evaluation rule and side effect.

"A"+new C()

following the same rule. It gets "A" first, which is a String literal, put it somewhere. Then it evaluate

new C() 

it construct a C Object first, then invoke toString() method of C Object, and gets the value of C object, which is "C", then concatenates "A" and "C" together, and println "AC".

Inside the toString() method of C Object, there is a System.out.print("B"); which is invoked when Java evaluate the above expression. It is printed out before the evaluation completed.
That is why "B" is printed first

Answer 3

Because the new C() is converted to a string, and then passed to println(). Basically, here's what happens:

1. Concatenate "A" with new C():
  a. Call String.valueOf(new C()):
    i. print "B"
    ii. return "C"
  b. Concatenate "A" and "C"
2. Pass "AC" to println
3. Print "AC"

AFAIK (I'm not 100% sure) string concatenation uses String#valueOf(Object) rather than directly calling Object#toString(). That's why "foo" + null is "foonull" rather than [throw a NullPointerException].