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
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
Let’s take a look at the intermediary language code for that: IL_0000: nop IL_0001: ldc.i4.s 2A IL_0003: stloc.0 // c IL_0004: ldloc.0 // c This loads the constant integer 42 onto the stack, then stores it into the variable c, and loads it immediately again onto the stack. IL_0005: stloc.1 IL_0006: ldloc.1 This copies the … Read more
Post-increment usually involves keeping a copy of the previous value around and adds a little extra code. Pre-increment simply does it’s job and gets out of the way. I typically pre-increment unless the semantics would change and post-increment is actually necessary.
You are right that it should not compile, and on most compilers, it does not compile. (Please specify exactly which compiler/version is NOT giving you a compiler error) I can only hypothesize that the compiler knows the identities that (i | i) == i and (i & i) == i and is using those identities … Read more
++i is slightly more efficient due to its semantics: ++i; // Fetch i, increment it, and return it i++; // Fetch i, copy it, increment i, return copy For int-like indices, the efficiency gain is minimal (if any). For iterators and other heavier-weight objects, avoiding that copy can be a real win (particularly if the … 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
As others have pointed out, the test is flawed in many ways. You did not tell us exactly how you did this test. However, I tried to implement a “naive” test (no offense) like this: class PrePostIncrement { public static void main(String args[]) { for (int j=0; j<3; j++) { for (int i=0; i<5; i++) … 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