operator-precedence

问题 From what I see operator precedence makes sense in these two examples: $a = false; $b = true; $c = $a || $b; Here $c is true $a = false; $b = true; $c = $a or $b; Here $c is false I understand the reasoning behind it. However the following: $a = false; $b = true; return $a or $b; Returns true, which puzzles me. What is the reason for this? 回答1: or has lower precedence than = , so this: $c = $a or $b; Becomes this: ($c = $a) or $b; But this doesn't make sense: (return $a) or $b; So you get

问题 I am having a hard time understanding some code that shows an example how a double in Java could be transformed into a byte[] and vice versa. Here is the code being used to transform a double into a byte[]: public static byte [] doubleToByteArray (double numDouble) { byte [] arrayByte = new byte [8]; long numLong; // Takes the double and sticks it into a long, without changing it numLong = Double.doubleToRawLongBits(numDouble); // Then we need to isolate each byte // The casting of byte (byte

问题 I am having a hard time understanding some code that shows an example how a double in Java could be transformed into a byte[] and vice versa. Here is the code being used to transform a double into a byte[]: public static byte [] doubleToByteArray (double numDouble) { byte [] arrayByte = new byte [8]; long numLong; // Takes the double and sticks it into a long, without changing it numLong = Double.doubleToRawLongBits(numDouble); // Then we need to isolate each byte // The casting of byte (byte

问题 I have this code, which is used to build a graphic interface on a LCD display on a controller; the code is compiled for 2 different architectures using both AVR and PIC32: FishinoTftGuiLabel *l1; FishinoTftGui .Page(F("Page1")) .Label(50, 140, 0, 24, LabelAlign::Left, F("Slider value:")) .getElement(l1) -- .Label(l1->x() + l1->w() + 10, 140, 0, 24, LabelAlign::Left, F("pippo")) ; Each member function return the same object (or a related one); so for example the Label() function returns the

问题 I have this code, which is used to build a graphic interface on a LCD display on a controller; the code is compiled for 2 different architectures using both AVR and PIC32: FishinoTftGuiLabel *l1; FishinoTftGui .Page(F("Page1")) .Label(50, 140, 0, 24, LabelAlign::Left, F("Slider value:")) .getElement(l1) -- .Label(l1->x() + l1->w() + 10, 140, 0, 24, LabelAlign::Left, F("pippo")) ; Each member function return the same object (or a related one); so for example the Label() function returns the

问题 There are lots of questions on concepts of precedence and order of evaluation but I failed to find one that refers to my special case. Consider the following statement: if(f(0) && g(0)) {}; Is it guaranteed that f(0) will be evaluated first? Notice that the operator is &&. My confusion stems from what I've read in "The C++ Programming Language, (Stroustrup, 4ed, 2013)". In section 10.3.2 of the book, it says: The order of evaluation of subexpressions within an expression is undefined . In

问题 This question already has answers here : How do chained comparisons in Python actually work? (1 answer) Python comparison operators chaining/grouping left to right? (1 answer) Closed 22 mins ago . Shouldn't it be parsed as (3 in [3]) == True since in and == have the same precedence? It's definitely not being parsed as 3 in ([3] == True) since that raises a TypeError . How else can it be parsed? p.s. I'm aware that PEP 8 says "Don't compare boolean values to True or False using == ." This is

问题 In the "Introduction" section of K&R C (2E) there is this paragraph: C, like any other language, has its blemishes. Some of the operators have the wrong precedence; ... Which operators are these? How are their precedence wrong? Is this one of these cases? 回答1: There is a clear rule of precedence that is incontrovertible. The rule is so clear that for a strongly typed system (think Pascal) the wrong precedence would give clear unambiguous syntax errors at compile time. The problem with C is

问题 I recently came across something that I thought I understood right off the bat, but thinking more on it I would like understanding on why it works the way it does. Consider the code below. The (x-- == 9) is clearly getting evaluated, while the (y++ == 11) is not. My first thought was that logical && kicks in, sees that the expression has already become false, and kicks out before evaluating the second part of the expression. The more I think about it, the more I don't understand why this

问题 In PHP, is if(!$foo) equivalent with if($foo != true) or with if($foo !== true) or is it even something completly different of both? 回答1: if(!$foo) is the equivalent to if($foo != true) so $foo = null; if(!$foo){ echo "asd"; } will ouptut "asd" 回答2: Note that, == OR != compares the values of variables for equality, type casting as necessary. === OR !== checks if the two variables are of the same type AND have the same value. This answer will give you better explanation of this concept: https: