syntax

问题 I'm working with this JS plugin, and I've encountered some syntax I've never seen before. I understand what it's doing, but I'm not sure why it works. Here's an example of one instance of it: settings.maxId != null && (params.max_id = settings.maxId); Is this just taking advantage of conditionals and the single = ? Is this common syntax for JS? 回答1: In JavaScript the = operator is an expression and evaluates the assigned value. Because it is an expression it can be used anywhere an expression

问题 I'm working with this JS plugin, and I've encountered some syntax I've never seen before. I understand what it's doing, but I'm not sure why it works. Here's an example of one instance of it: settings.maxId != null && (params.max_id = settings.maxId); Is this just taking advantage of conditionals and the single = ? Is this common syntax for JS? 回答1: In JavaScript the = operator is an expression and evaluates the assigned value. Because it is an expression it can be used anywhere an expression

问题 I'm working with this JS plugin, and I've encountered some syntax I've never seen before. I understand what it's doing, but I'm not sure why it works. Here's an example of one instance of it: settings.maxId != null && (params.max_id = settings.maxId); Is this just taking advantage of conditionals and the single = ? Is this common syntax for JS? 回答1: In JavaScript the = operator is an expression and evaluates the assigned value. Because it is an expression it can be used anywhere an expression

问题 If it can be initiated with just String s = "Hello"; then why is it a class? Where's the parameters? 回答1: Given that String is such a useful and frequently used class, it has a special syntax (via a string literal representation: the text inside "" ) for creating its instances, but semantically these two are equivalent: String s = "Hello"; // just syntactic sugar String s = new String("Hello"); Behind the hood both forms are not 100% equivalent, as the syntax using "" tries to reuse strings

问题 This question already has answers here : Normal arguments vs. keyword arguments (10 answers) Closed 6 years ago . This is an example of of code from here: What does the equality mean in the argument assignment to function? like N=20000 here? What is the difference between that and simply N as argument? import random,math def gibbs(N=20000,thin=500): x=0 y=0 samples = [] for i in range(N): for j in range(thin): x=random.gammavariate(3,1.0/(y*y+4)) y=random.gauss(1.0/(x+1),1.0/math.sqrt(x+1))

问题 I was playing around with some code and I was wondering if any can tell me what the curly braces in this code represents. I thought it would've been for an empty object but that doesn't seem to be the case. Person person = new Person{}; if (person is {}){ Console.WriteLine("Person is empty."); } else { Console.WriteLine("Person is not empty."); } It compiles just fine; but if I populate the properties of the person class it still falls into the person is empty part of the if statement. 回答1: {

问题 Looking at this page: http://www.mikeash.com/pyblog/friday-qa-2010-12-31-c-macro-tips-and-tricks.html I found this snippet of code with ^{ ... }() syntax, what are the caret/brackets doing? #define MAX(x, y) (^{ \ int my_localx = (x); \ int my_localy = (y); \ return my_localx > my_localy ? (my_localx) : (my_localy); \ }()) It looks like its creating an anonymous function or something. What is this concept called? Where can I read about it? 回答1: It's a C block. It's quite like an anonymous

问题 Looking at this page: http://www.mikeash.com/pyblog/friday-qa-2010-12-31-c-macro-tips-and-tricks.html I found this snippet of code with ^{ ... }() syntax, what are the caret/brackets doing? #define MAX(x, y) (^{ \ int my_localx = (x); \ int my_localy = (y); \ return my_localx > my_localy ? (my_localx) : (my_localy); \ }()) It looks like its creating an anonymous function or something. What is this concept called? Where can I read about it? 回答1: It's a C block. It's quite like an anonymous

问题 I'm having a syntax error with this if block and can't I haven't been able to correct it if [[ $X >= 100] || [$Y >= 100 ]] then echo "..." fi I've rewrote this, but haven't had any luck on finding the correct syntax. Thanks in advance! 回答1: This is a syntax error, you should try : if ((X >= 100 || Y >= 100 )) then echo "..." fi NOTE with this syntax, no need to remember -ge and such. This is just like arithmetic ((...)) is an arithmetic command, which returns an exit status of 0 if the

问题 I am upgrading my preexisting CAD/CAM project (quite big one > 10MByte of code) and have to add some special measuring equipment. The problem is I have more than one supplier of the measuring system (although is already decided which one to use) and I want to configure them (using #define in case vendor is changed in future) with code to use only selected device type. So I have something like: #define use_vendor1 //#define use_vendor2 //#define use_vendor3 and some of the vendors APIs require