Java When outputting String and method return, why does method return output first?

Question

In the code below, if the string \"Mult\" comes before the test1(4) method call, why does the method output before the string? And why does it bounce f

Answer 1

The first thing to note is when you use the + with two operands where one of the two operands is a String, the result of the expression is a String.

Therefore, in the following method invocation expression

System.out.println("Mult:" + test1(4));

you are invoking the PrintStream#println(String) since out is a variable of type PrintStream. Note how the method accepts a single String argument. Therefore, the String has to be resolved from the String concatenation of

"Mult:" + test1(4)

For that to happen, the test1(4) method has to be executed.

public static int test1(int n){
    System.out.println("N:" + n);
    return n*2;
}

This method again uses PrintStream#println(String) but with the argument

"N:" + n

This is another String concatenation that produces the String value

"N:4"

for this particular invocation. The produced String value is then used as an argument to the println(..) method which outputs it to your program's standard output.

The method then returns the value 8, since 4 * 2 = 8.

That return value is the value of invoking the test1(4) method. So

System.out.println("Mult:" + test1(4));

is equivalent to

System.out.println("Mult:" + 8);

Then String concatenation occurs, transforming

"Mult:" + 8

into the String value

"Mult:8"

That String is then used as the single argument to the println(..) method which outputs it to your program's standard output.