operators

问题 In an example of Objective-C code I found this operator self.itemViews[@(0)] ?: [self.dataSource slidingViewStack:self viewForItemAtIndex:0 reusingView:[self dequeueItemView]]; The code does compile under Apple LLVM 4.2. The only thing I came across was in being a vector operator, but I don't think Objective-C, and for that matter C, has vector operators. So can someone please give reference and or documentation of this operator. 回答1: ?: is the C conditional operator . a ? b : c yields b

问题 I know this code will work: echo ( $a == $b || $a == $c ) ? "Yes" : "No"; That can be read like: if $a is equal to $b or $a is equal to $c Is there a way to make it more shorter like: if $a is equal to $b or $c I have tried a lot including this but still no luck: echo ( $a == ( $b xor $c ) ) ? "Yes" : "No"; 回答1: You can use in_array: var_dump(in_array($a, [$b, $c])); with your example: echo in_array($a, [$b, $c]) ? 'Yes' : 'No'; Note: this syntax is only useful if you have more than 2 values.

问题 Error: error C2678: binary '==' : no operator found which takes a left-hand operand of type 'const entry' (or there is no acceptable conversion) The function: template <class T, int maxSize> int indexList<T, maxSize>::search(const T& target) const { for (int i = 0; i < maxSize; i++) if (elements[i] == target) //ERROR??? return i; // target found at position i // target not found return -1; } indexList.h indexList.cpp Is this suppose to be an overloaded operator? Being a template class I am

问题 Is it possible to define a brand new operator in Groovy? I would like to express a trade where someone buys 200 items for the price of 10 like this: def trade = 200 @ 10 Is this achievable? Thanks EDIT: I want to make it clearer that I am interested in defining an operator not adding a method. Cheers. 回答1: I am not quite sure how you can make this work for the @ sign but you could certainly add the operation like this which I actually find more expressive: Number.metaClass.buyFor { Integer

问题 I've found in 'addons/math/misc/brent.ijs' implementation of Brent's method as an adverb. I would like to build a Newton's method as an adverb too but it's much harder than building tacit verbs. Here is a explicit version of Newton's iteration: newton_i =: 1 : '] - u % u d.1' With such usage: 2&o. newton_i^:_ (1) NB. (-: 1p1) must be found 1.5708 2 o. 1.5708 NB. after substitution we get almost 0 _3.67321e_6 And of course, for convenience: newton =: 1 : 'u newton_i^:_' What's a tacit

问题 What's the name of this operator "+=" ? 回答1: It, along with -= , *= , etc., are called the augmented assignment operators in Python, and "compound assignment" operators everywhere else. 回答2: The name is "plus equal" operator! 回答3: In c# it's called the addition assignment operator. 回答4: That is the Addition Assignment operator. 回答5: += is the plus and Equal operator. If you assign a+=3 That means u assign the expression are a=a+3. 来源： https://stackoverflow.com/questions/2107333/whats-the-name

问题 In any programming language textbooks, we are always told how each operator in that language has either left or right associativity. It seems that associativity is a fundamental property of any operator regardless of the number of operands it takes. It also seems to me that we can assign any associativity to any operator regardless of how we assign associativity to other operators. But why is it the case? Perhaps an example is better. Suppose I want to design a hypothetical programming

问题 In any programming language textbooks, we are always told how each operator in that language has either left or right associativity. It seems that associativity is a fundamental property of any operator regardless of the number of operands it takes. It also seems to me that we can assign any associativity to any operator regardless of how we assign associativity to other operators. But why is it the case? Perhaps an example is better. Suppose I want to design a hypothetical programming

问题 I saw if($output !== false){ } It's an exclamation mark with two equals signs. It almost works like not equal. Does it has any extra significance? 回答1: They are the strict equality operators ( ===, !==) , the two operands must have the same type and value in order the result to be true. For example: var_dump(0 == "0"); // true var_dump("1" == "01"); // true var_dump("1" == true); // true var_dump(0 === "0"); // false var_dump("1" === "01"); // false var_dump("1" === true); // false More

问题 I found this operator by chance: ruby-1.9.2-p290 :028 > "abc" !=~ /abc/ => true what's this? It's behavior doesn't look like "not match". 回答1: That's not one operator, that's two operators written to look like one operator. From the operator precedence table (highest to lowest): [] []= ** ! ~ + - [unary] [several more lines] <=> == === != =~ !~ Also, the Regexp class has a unary ~ operator: ~ rxp → integer or nil Match—Matches rxp against the contents of $_ . Equivalent to rxp =~ $_ . So your