Note: The two defect reports DR#637 and DR#222 are important to understand the below’s behavior rationale. For explanation, in C++0x there are value computations and side effects. A side effect for example is an assigment, and a value computation is determining what an lvalue refers to or reading the value out of an lvalue. Note … 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
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
Executive summary: No. i++ could potentially be slower than ++i, since the old value of i might need to be saved for later use, but in practice all modern compilers will optimize this away. We can demonstrate this by looking at the code for this function, both with ++i and i++. $ cat i++.c extern … Read more
i = 5; System.out.println(++i); //6 This prints out “6” because it takes i, adds one to it, and returns the value: 5+1=6. This is prefixing, adding to the number before using it in the operation. i = 6; System.out.println(i++); //6 (i = 7, prints 6) This prints out “6” because it takes i, stores a … Read more
All four answers so far are incorrect, in that they assert a specific order of events. Believing that “urban legend” has led many a novice (and professional) astray, to wit, the endless stream of questions about Undefined Behavior in expressions. So. For the built-in C++ prefix operator, ++x increments x and produces (as the expression’s … Read more