Usage and type conversion of JAVA three-mesh operator Boolean?condition1:condition2 __java

Tri-Mesh operator < expression 1>?< expression 2>:< expression 3>; "?" The meaning of an operator is to first evaluate the value of expression 1 and, if true, execute Expression 2 and return the result of expression 2; If the value of expression 1 is false, the expression 3 is executed and the result of expression 3 is returned.

The above is the basic definition and use of the three-mesh operator. On the surface, it should be relatively simple. In the Book of Java Programmer interview, we have seen two interesting topics.

Topic 1: "China northeast famous software company D2009 March test"

int a=5;
System.out.println ("a=" + (a<5) (10.9:9));

A. Compilation errors

B. 10.9

C. 9

D. None of the above answers are correct

Perhaps like most people, just start to assume that A<5 is false, then the result is 9, choose c.

Think carefully, this is the problem set trap. In the expression = (a<5)? 10.9:9 There is a 10.9, this is the Java will be based on the operator's accuracy of automatic type conversion, because the front is 10.9, then the following 9 will then become 9.0!

Topic 2: "China Northeast famous software company D2009 March test"

Char x= ' x ';
int i=10;
System.out.println (FALSE?I:X);
System.out.println (FALSE?100:X);

A. x

B. 120 120

C. x 120

D. None of the above answers are correct

The answer is a.

Parsing: System.out.println (false?i:x) is the same as the last topic 1, X is promoted to int type, so output X's ASCII code
For the second row, because 100 is a constant expression. If the two expressions in the three-mesh operator are a constant expression and the other is an expression of type T (char in this case), and the constant expression can be represented by T, the output result is the type T. So the output is a character

Core ideas:

1. If one of the two expressions in the three-mesh operator is a constant expression and the other is an expression of type T, and a constant expression can be represented by T, the output result is the type T.

2. If all are constant expressions, use the UP type conversion

Enter the exercise mode:

(1)

Char x= ' x ';
int i=10;

System.out.println (FALSE?10000:X);

System output: X

Okay, what if it's modified?

Char x= ' x ';
int i=10;

System.out.println (FALSE?100000:X);

What is the output of the system? 120.

Parsing: "Core idea 1" 10000 can be represented by the T type of the subject, so the output is T type, output x, but 100000 is not, because it exceeds the ASCII range. Then we're going to output the ASCII value of X.

Continue to extend

Char x= ' x ';

System.out.println (TRUE?100:X);

What about this one.

Resolution: The same applies to "core idea 1", after true enter condition1.100 can be represented by T type, then input will be T type (char type), 100 corresponds to ASCII is D, then output d.

(2)

Char x= ' x ';

System.out.println (TRUE?X:10);

System.out.println (true?x:10.0);

OK, the first line should output is X, this is no problem. So the second line. Results: 120.0

Parsing: Apply "core idea 1", but 10.0 cannot be type T, then output will not be T type. 120 will be upgraded to double type to 120.0

(3)

System.out.println (fasle?9:10.0);

System.out.println (true?9:10.0);

Results: 10.0 9.0

Resolution: For Core idea 2, the result of the first line Fasle is the highest type, and the second line 9 will be upgraded to double to 9.0