operator-precedence

问题 I have overloaded some Python operators, arithmetic and boolean. The Python precedence rules remain in effect, which is unnatural for the overloaded operators, leading to lots of parentheses in expressions. Is there a way to "overload" Python's precedences? 回答1: You can cheat that mechanism in this way: Override all operators to not do the calculations but create list of instructions wrapped in some object. Add your own bracket operator (ie. as a _ function). Example: >>> a = MyNumber(5); b =

问题 Will the differences below matter significantly in C#? int a, b; double result; result = (double)a / b; result = a / (double)b; result = (double)a / (double)b; Which one do you use? 回答1: The cast will occur before the division. In your examples, it doesn't matter which one you do as if one operand is a double, the runtime will cast/convert the other to a double as well. This looks like a micro-optimization - not something worth worrying about or dealing with unless measurements show it is

问题 In previous versions of the standard (C++03) the order of evaluation of parameters to a function call was unspecified. Has this been changed in subsequent version of the standard (C++11 or C++14)? i.e. Can we rely on a specific ordering (left to right) or not. 回答1: No this has not changed but there is a very recent proposal to change this: N4228: Refining Expression Evaluation Order for Idiomatic C++, this was part of the Pre-Urbana mailing that came out this October The introduction says (

问题 int main() { int i=3; (++i)++; printf("%d",i); } This programs works with g++ compiler but not gcc. If i write i++++ or ++i++ it doesn't work in cpp also. I think there is difference between c-expression and c++-expression. Can somebody explain about L-value and R-value ? 回答1: Edit: This answer is incorrect for the updated question, it applies to the question as originally stated. (i++)++ shouldn't work with either gcc or g++, whether or not they are parsing the file as C or C++ in both

问题 I've searched the web but I've found no solution to this problem. What is the logical priority for operators NAND , NOR and XNOR ? I mean, considering as example the expression A AND B NAND C which operator should be evaluated first? Obviously NAND can be translated as NOT-AND (as NOR is NOT-OR and XNOR is NOT-XOR ), but (A AND B) NAND C != A AND (B NAND C) = A AND NOT(B AND C) According to my researches there's no a defined priority for such an expression, so I think the simplest solution is

问题 There's been a change in Swift 3 for Xcode 8 beta 6 and now I'm not able to declare my operator for power as I did before: infix operator ^^ { } public func ^^ (radix: Double, power: Double) -> Double { return pow((radix), (power)) } I've read a bit about it and there's a new change been introduced in Xcode 8 beta 6 From this I'm guessing I have to declare a precedence group and use it for my operator like this: precedencegroup ExponentiativePrecedence {} infix operator ^^:

问题 Say I have this code: unsigned int func1(); unsigned int func2(); unsigned int func3(); unsigned int x = func1() | func2() | func3(); Does C++ guarantee that func1() will be called first, then func2(), and then func3()? Or is the compiler allowed to call the functions in any order it feels like? Also, is the compiler allowed to implement a short-circuit optimization here if it wants to? (e.g. if func1() returned ~0, could the compiler decide not to bother calling func2() or func3(), because

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 4 years ago . Can some one tell why my php line break not working ( echoing ) ? I know i can write the code in a different way to make the line break work, but i want to know the reason behind this ? <?php $var1 = 3; echo "Addition = " . $var1 += 3 . "<br>"; echo "Subtraction = " . $var1 -= 3 . "<br>"; echo

问题 If I run the following code on Firefox I get an error: new Number.toString; But according to MDN https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Operator_Precedence new Number should evaluate first. So the table is not correct I think. Let us take a look at MSDN: http://msdn.microsoft.com/en-us/library/z3ks45k7(v=vs.94).aspx . Above the table is written that operators are evaluated from left to right. But: a=1; b=a=2; Now b has the value 2 which suggest evaluation

问题 The very simple Java code as follows has the weird output, but the same logic code in C and C++ has the right output. I try with the JDK 1.7 and JDK 1.3 (relative JRE), the weird output is always there. public class Test { public static int sum=0; public static int fun(int n) { if (n == 1) return 1; else sum += fun(n - 1); // this statement leads to weird output // { // the following block has right output // int tmp = fun(n - 1); // sum += tmp; // } return sum; } public static void main