operator-keyword | 易学教程

Operator precedence in Python -PEMDAS

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问题 I read about python following PEMDAS that is precedence of multiply is more than division. I ran the following script print 6*2/1*2 Thus python should interpret this like 12/2 i.e 6 , since precedence of multiplication is more than division. But, the answer is 24. Could anyone let me know where the problem is? Thanks! 回答1: * has the same operator precedence as / . Operators in the same group evaluate left to right, so your expression evaluates as: 6*2 = 12 / 1 = 12 * 2 = 24 回答2: Order of

What is the Python <> operator

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问题 What exactly is the <> operator in Python, and why is it undocumented (as far as I can tell)? Is it the same as != or is not ? 回答1: In Python 2.x, <> is the same as != (i.e. "not equal to" , rather than is not which is "not identical to" ), but the latter is preferred: The comparison operators <> and != are alternate spellings of the same operator. != is the preferred spelling; <> is obsolescent. In 3.x, <> has been removed and only != exists. 回答2: It is documented, but you're not supposed to

operator<< overloading ostream

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问题 In order to use cout as such : std::cout << myObject, why do I have to pass an ostream object? I thought that was an implicit parameter. ostream &operator<<(ostream &out, const myClass &o) { out << o.fname << " " << o.lname; return out; } Thanks 回答1: You aren't adding another member function to ostream , since that would require redefining the class. You can't add it to myClass , since the ostream goes first. The only thing you can do is add an overload to an independent function, which is

mouse-over to peek a field after operator->() in Visual Studio while debugging

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问题 I had a tiny class:- class A{ public:int aField; } Below, while debugging, if I hover mouse around aField in a->aField , Visual Studio will pop up the value of the field nicely (like a tiny Watch). A* a=new A(); a->aField=1234; //^ hover here Then I upgraded code to override operator-> :- class APtr{ //my custom smart pointer A* ptr; A* operator->(){ return ptr; } } APtr a; ..... a->aField=1234; //^ hover here There is no pop up anymore. (There is a popup for a , but not for aField ) How to

ternary operator without else in C

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I want to use ternary operator without else in C. How do I do it. (a)? b: nothing; something like this. What do I use in nothing part? If you are using a ternary operator like that, presumably it could be replaced by: if (a) { b; } which is much, much better. (The intent is clearer, so the code is easier to read, and there will be no performance loss.) However, if you are using the ternary operator as an expression, i.e. printf("%d cat%s", number_of_cats, number_of_cats != 1 ? "s" : <nothing>); a = b*c + (d == 0 ? 1 : <nothing>); then the <nothing> value depends on the context it is being used

String concatenation does not work properly in Java when concatenating 2 results of ternary operators

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问题 Dear Java guru 's! Can you, please, explain me, why String concatenation does not work properly in Java when concatenating 2 results of ternary operators? Example: String str = null; String x = str != null ? "A" : "B" + str == null ? "C" : "D"; System.out.println(x); Output is "D", but I expected "BC". I am suspecting that it works like so because of operators priorities, but I am not sure, about how we exactly we get "D" for case above. What calculation algorithm takes place for this case?

' << ' operator in verilog

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i have a verilog code in which there is a line as follows: parameter ADDR_WIDTH = 8 ; parameter RAM_DEPTH = 1 << ADDR_WIDTH; here what will be stored in RAM_DEPTH and what does the << operator do here. << is a binary shift, shifting 1 to the left 8 places. 4'b0001 << 1 => 4'b0010 >> is a binary right shift adding 0's to the MSB. >>> is a signed shift which maintains the value of the MSB if the left input is signed. 4'sb1011 >> 1 => 0101 4'sb1011 >>> 1 => 1101 Three ways to indicate left operand is signed: module shift; logic [3:0] test1 = 4'b1000; logic signed [3:0] test2 = 4'b1000; initial

The operator is undefined

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I just tried to make a simple class that lets me figure out the length of a file: public class Size { long s = 0; int a; public static void main(String[]args){ new Size(); } Size(){ try{ FileInputStream str = new FileInputStream("E:/Eclipse/Resources/smile.jpg"); while(a != null){ s++; } }catch (IOException e){ e.printStackTrace(); } } } I run into a problem with while(a != null) I get the Error: The operator != is undefined for the argument type(s) int, null Any ideas why it's blocking the condition? sonOfRa Primitive types in Java cannot be null . If you want to check for 0, do a != 0 . Put

Hack to convert javascript number to UInt32

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Edit: This question is out of date as the Polyfill example has been updated. I'm leaving the question here just for reference. Read the correct answer for useful information on bitwise shift operators. Question: On line 7 in the Polyfill example of the Mozilla Array.prototype.indexOf page they comment this: var length = this.length >>> 0; // Hack to convert object.length to a UInt32 But the bitwise shift specification on Mozilla clearly states that the operator returns a value of the same type as the left operand: Shift operators convert their operands to thirty-two-bit integers and return a

C++ overload operator twice, one return non-const reference and the other const reference, what is the preference?

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问题 I overload an operator twice with the same parameter list. but with different return type: T& operator()(par_list){blablabla} const T& operator()(par_list){blablabla} So when I'm calling the () operator, which function would be called based on what preference or situation? I know that if I call () under const function it has to be the const T& one. I'm just curious how C++ deal with such situation and how the default preference works. Thanks 回答1: These functions don't overload each other;