operator-precedence

If I have a piece of code written in C# wrapped in an #if directive, what (if any) precedence is applied to any boolean operators that might be used in that directive? In other words: #if DEBUG || MYTEST && PLATFORM_WINDOWS // ... Some code here #endif Will that be simply evaluated left to right as #if (DEBUG || MYTEST) && PLATFORM_WINDOWS And similarly, would #if PLATFORM_WINDOWS && DEBUG || MYTEST Be evaluated as #if (PLATFORM_WINDOWS && DEBUG) || MYTEST Or is there some precedence order for && vs ||? Edit: To be clear, I am well aware that I can run the code myself to test it, and I have. I

Let's say we have class A : class A { public: A& func1( int ) { return *this; } A& func2( int ) { return *this; } }; and 2 standlone functions: int func3(); int func4(); now in this code: A a; a.func1( func3() ).func2( func4() ); is order of evaluation of functions func3() and func4() defined? According to this answer Undefined behavior and sequence points one of the sequence points are: at a function call (whether or not the function is inline), after the evaluation of all function arguments (if any) which takes place before execution of any expressions or statements in the function body ( §1

If I give Mathematica the input TreeForm[Unevaluated[4^5]] I expect to see three boxes -- power, 4, and 5. Instead I see a single box with 1024. Can anyone explain? Compare TreeForm@Unevaluated[4^5] with TreeForm@Hold[4^5] From the help: Unevaluated[expr] represents the unevaluated form of expr when it appears as the argument to a function. and Hold[expr] maintains expr in an unevaluated form. so, as Unevaluated[4^5] gets to TreeForm ... it gets evaluated ... It works like this: f[x_+y_]:=x^y; f[3+4] (* -> f[7] *) f[Unevaluated[3+4]] (* ->81 *) A level of Unevaluated is stripped off with every

What is the correct sequence of the math operations in this expression in Java: a + b * c / ( d - e ) 1. 4 1 3 2 2. 4 2 3 1 I understand that result is the same in both answers. But I would like to fully understand the java compiler logic. What is executed first in this example - multiplication or the expression in parentheses? A link to the documentation that covers that would be helpful. UPDATE: Thank you guys for the answers. Most of you write that the expression in parentheses is evaluated first. After looking at the references provided by Grodriguez I created little tests: int i = 2;

Note: No multithreading at all here. Just optimized single-threaded code. A function call introduces a sequence point . (Apparently.) Does it follow that a compiler (if the optimizer inlines the function) is not allowed to move/intermingle any instructions prior/after with the function's instructions? (As long as it can "proove" no observable effects obviously.) Explanatory background: Now, there is a nice article wrt. a benchmarking class for C++, where the author stated: The code we time won’t be rearranged by the optimizer and will always lie between those start / end calls to now(), so we

Possible Duplicates: Output of multiple post and pre increments in one statement Post-increment and pre-increment in 'for' loop The following code snippet int i=0; printf("%d %d",i++,i++); gives the output 1 0 I can understand that, but the following int i=0; printf("%d %d",++i,++i); gives the output 2 2 Can someone explain me the second behavior? Nawaz Both printfs invoke undefined-behavior. See this : Undefined behavior and sequence points Quoted from this link: In short, undefined behaviour means anything can happen from daemons flying out of your nose to your girlfriend getting pregnant.

#include <iostream> int foo() { std::cout<<"foo() is called\n"; return 9; } int bar() { std::cout<<"bar() is called\n"; return 18; } int main() { std::cout<<foo()<<' '<<bar()<<' '<<'\n'; } // Above program's behaviour is unspecified // clang++ evaluates function arguments from left to right: http://melpon.org/wandbox/permlink/STnvMm1YVrrSRSsB // g++ & MSVC++ evaluates function arguments from right to left // so either foo() or bar() can be called first depending upon compiler. Output of above program is compiler dependent. Order in which function arguments are evaluated is unspecified . The

I have an idea for a simple program to make that will help me with operator precedence in languages like C. The most difficult part of this is parenthesizing the expression. For example, I want this: *a.x++ = *b.x++ Converted to this: ((*(((a).(x))++)) = (*(((b).(x))++))) Which I did manually in these steps: *a.x++ = *b.x++ *(a).(x)++ = *(b).(x)++ *((a).(x))++ = *((b).(x))++ *(((a).(x))++) = *(((b).(x))++) (*(((a).(x))++)) = (*(((b).(x))++)) ((*(((a).(x))++)) = (*(((b).(x))++))) What is the best way to accomplish this? Is there already a solution out there that I could use? I'd prefer to do