The loop is equivalent to: int x = 2; { int y = 2; while (y > 0) { System.out.println(x + ” “+ y + ” “); x++; y–; // or –y; } } As you can see from reading that code, it doesn’t matter whether you use the post or pre decrement operator in … Read more
Part-1: Array names are constant (not modifiable lvalue), your can add value to array name but can’t modify it. Expression a + 2 doesn’t modify a itself but when you do a++ that is equivalent to a = a + 1 try to modify array name –lvalue error. The expression a++ in second printf is … Read more
You hit the nail on the head. Your understanding is correct. The difference between pre and post increment expressions is just like it sounds. Pre-incrementation means the variable is incremented before the expression is set or evaluated. Post-incrementation means the expression is set or evaluated, and then the variable is altered. It’s easy to think … Read more
Quoting Java Language Specification, 15.7 Evaluation Order: The Java programming language guarantees that the operands of operators appear to be evaluated in a specific evaluation order, namely, from left to right. The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated. If the operator … Read more
x– will be 4, but will be 3 at the moment of –x, so it will end being 2, then you’ll have x = 4 – 2 btw, your first case will be x = 4 + 6 Here is a small example that will print out the values for each part, maybe this way … Read more
Well, there are two things to consider with your example code: The order of evaluation of function arguments is unspecified, so whether ++a or a++ is evaluated first is implementation-dependent. Modifying the value of a more than once without a sequence point in between the modifications results in undefined behavior. So, the results of your … Read more
i++ : create a temporary copy of i increment i return the temporary copy ++i : increment i return i With optimizations on, it is quite possible that the resulting assembly is identical, however ++i is more efficient. edit : keep in mind that in C++, i may be whatever object that support the prefix … Read more
Since the statement i++ ends at the ; in your example, it makes no difference whether you use pre- or post-increment. The difference arises when you utilize the result: int j = i++; // i will contain i_old + 1, j will contain the i_old. Vs: int j = ++i; // i and j will … Read more
Let’s break down your own argument: According to the author, j=j++; is similar to temp=j; j=j+1; // increment j=temp; // then assign Yes, you’re right so far…, but here’s where you got it wrong: But a=b++; makes b=1. So it should’ve evaluated like this, a=b; // assign b=b+1; // then increment WRONG! You’re not applying … Read more
It’s not a question of preference, but of logic. x++ increments the value of variable x after processing the current statement. ++x increments the value of variable x before processing the current statement. So just decide on the logic you write. x += ++i will increment i and add i+1 to x. x += i++ … Read more