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Okay, so I have this piece of code tested and I found there isn't any exception thrown out.
public static void main(String[] args) { int[] list = {1,2}; if (list.length>2 && list[3] == 2){ System.out.println(list[1]); } } Does the statement here
if (list.length>2 && list[3] == 2) mean that if the first condition is false we don't even have to check the second condition?
Or it equals to
if (list.length>2){ if (list[3] == 2){ ... } } ?
And, what if it is written in C or python or some other languages?
Thanks
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Yes, the order of the conditions matters. In your code, you test if(a==b) first. If all 3 integers are the same, then this will be true and only return c; will execute. It won't even bother with the rest of the conditions.
This is called lazy evaluation. In order to make it greedy, use the single ampersand logical operator. In that case, it will also evaluate the second conditional, even if the first turns out to be false.
Use two if statements if both if statement conditions could be true at the same time. In this example, both conditions can be true. You can pass and do great at the same time. Use an if/else statement if the two conditions are mutually exclusive meaning if one condition is true the other condition must be false.
If the conditional is not true then the line of code immediately after the else statement will be executed and then "flow of control" will pass to the next statement following the if-else statement.
It is common for languages (Java and Python are among them) to evaluate the first argument of a logical AND and finish evaluation of the statement if the first argument is false. This is because:
From The Order of Evaluation of Logic Operators,
When Java evaluates the expression d = b && c;, it first checks whether b is true. Here b is false, so b && c must be false regardless of whether c is or is not true, so Java doesn't bother checking the value of c.
This is known as short-circuit evaluation, and is also referred to in the Java docs.
It is common to see list.count > 0 && list[0] == "Something" to check a list element, if it exists.
It is also worth mentioning that if (list.length>2 && list[3] == 2) is not equal to the second case
if (list.length>2){ if (list[3] == 2){ ... } } if there is an else afterwards. The else will apply only to the if statement to which it is attached.
To demonstrate this gotcha:
if (x.kind.equals("Human")) { if (x.name.equals("Jordan")) { System.out.println("Hello Jordan!"); } } else { System.out.println("You are not a human!"); } will work as expected, but
if (x.kind.equals("Human") && x.name.equals("Jordan")) { System.out.println("Hello Jordan!"); } else { System.out.println("You are not a human!"); } will also tell any Human who isn't Jordan they are not human.
60
Yes. If the first condition is not satisfied then the remainder are not evaluated. This is known as short-circuiting. See here for more details. Note that this isn't particular to Java and lot of other languages will do the same.
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