pre-increment

During a programming class, the professor was teaching us about x++ and ++x , with x being an integer. He said that in the scenario we are able to just put either x++ or ++x , ++x is more efficient (by little, but still, in theory, more efficient nonetheless). But I forgot why . Anyone knows? This was with Java. zneak It's not more efficient in Java. It can be more efficient in languages where the increment/decrement operators can be overloaded, but otherwise the performance is exactly the same. The difference between x++ and ++x is that x++ returns the value of x before it was incremented,

I'm making a program in java that races a few cars against each other. Each car is a separate thread. When cars complete the race, each one all calls this method. I've tested the method at varying timer speeds, and it seems to work fine. But I do realize that each thread is accessing the variable carsComplete, sometimes at the exact same time (at least at the scope the date command is giving me). So my question is: is this method thread-safe? public static String completeRace() { Date accessDate = new Date(); System.out.println("Cars Complete: " + carsComplete + " Accessed at " + accessDate

I was expecting that in my following code: #include<stdio.h> int main(){ int i = 10; int j = 10; j = ++(i | i); printf("%d %d\n", j, i); j = ++(i & i); printf("%d %d\n", j, i); return 1; } expressions j = ++(i | i); and j = ++(i & i); will produce lvalue errors as below: x.c: In function ‘main’: x.c:6: error: lvalue required as increment operand x.c:9: error: lvalue required as increment operand But I surprised that above code compiled successfully, as below: ~$ gcc x.c -Wall ~$ ./a.out 11 11 12 12 Check the above code working correctly. While other operators produce error (as I understand).

I am confused about the post ++ and pre ++ operator , for example in the following code int x = 10; x = x++; sysout(x); will print 10 ? It prints 10,but I expected it should print 11 but when I do x = ++x; instead of x = x++; it will print eleven as I expected , so why does x = x++; doesn't change the the value of x ? No, the printout of 10 is correct. The key to understanding the reason behind the result is the difference between pre-increment ++x and post-increment x++ compound assignments. When you use pre-increment, the value of the expression is taken after performing the increment. When

This question already has an answer here: How do the post increment (i++) and pre increment (++i) operators work in Java? 14 answers In programming, particularly in Java, what is the difference between: int var = 0; var++; and int var = 0; ++var; What repercussions would this have on a for loop? e.g. for (int i = 0; i < 10; i++) {} for (int i = 0; i < 10; ++i) {} Dónal tldr; Although both var++ and ++var increment the variable they are applied to, the result returned by var++ is the value of the variable before incrementing, whereas the result returned by ++var is the value of the variable

I am a little confused about how the C# compiler handles pre- and post increments and decrements. When I code the following: int x = 4; x = x++ + ++x; x will have the value 10 afterwards. I think this is because the pre-increment sets x to 5 , which makes it 5+5 which evaluates to 10 . Then the post-increment will update x to 6 , but this value will not be used because then 10 will be assigned to x . But when I code: int x = 4; x = x-- - --x; then x will be 2 afterwards. Can anyone explain why this is the case? Sebastian Piu x-- will be 4, but will be 3 at the moment of --x , so it will end

Why does the following compile in C++? int phew = 53; ++++++++++phew ; The same code fails in C, why? Prasoon Saurav That is because in C++ pre-increment operator returns an lvalue and it requires its operand to be an lvalue . ++++++++++phew ; in interpreted as ++(++(++(++(++phew)))) However your code invokes Undefined Behaviour because you are trying to modify the value of phew more than once between two sequence points . In C , pre-increment operator returns an rvalue and requires its operand to be an lvalue . So your code doesn't compile in C mode. Note: The two defect reports DR#637 and DR