tostring

Are JavaScript template literals guaranteed to call toString()?

由 陌路散爱 提交于 2020-11-24 18:07:30
问题 const num = 42 const str = `My number is ${num}` In this code what guarantee do I have about the conversion of num to a string ? Is it guaranteed to just call its toString() method or could the conversion be done in another way ? 回答1: Untagged templates use the ECMAScript ToString() abstract operation. The logic of template literal evaluation is spread over several sections which makes it difficult to follow, so I'll just post a link to it: https://tc39.es/ecma262/#sec-template-literals

Objects.deepToString(Object o) method

由 ⅰ亾dé卋堺 提交于 2020-07-04 01:58:32
问题 The class java.util.Objects contains the deepEquals(Object a, Object b) method that can be used to compare objects of any type (including arrays and null references), but doesn't contain a similar deepToString(Object o) . This is disappointing. (By the way, the private constructor of this class contains the message "No java.util.Objects instances for you!" that explains to some extent why this class is so mean). That being the case, I've tried to implement the method myself: public static

Making a user-defined class std::to_string(able)

由 十年热恋 提交于 2020-06-24 04:12:08
问题 I know it seems too much Java or C#. However, is it possible/good/wise to make my own class valid as an input for the function std::to_string ? Example: class my_class{ public: std::string give_me_a_string_of_you() const{ return "I am " + std::to_string(i); } int i; }; void main(){ my_class my_object; std::cout<< std::to_string(my_object); } If there is no such thing (and I think that), what is the best way to do it? 回答1: First, some ADL helping: namespace notstd { namespace adl_helper {

toString method for inherited case class in Scala

由 ∥☆過路亽．° 提交于 2020-05-16 06:33:32
问题 I am facing some inconsistency in calling toString method for case-classes in Scala. The first code sample: case class Person(name: String, age: Int) val jim = Person("jim", 42) println(jim) output: Person(jim,42) For the next code sample I used a case class that extends Exception : case class JimOverslept(msg: String) extends Exception try { throw JimOverslept(msg = "went to bed late") } catch { case e: JimOverslept => println(e) } output: playground.CaseClassOutput$JimOverslept Actually, I

toString method for inherited case class in Scala

由 只愿长相守 提交于 2020-05-16 06:33:19
问题 I am facing some inconsistency in calling toString method for case-classes in Scala. The first code sample: case class Person(name: String, age: Int) val jim = Person("jim", 42) println(jim) output: Person(jim,42) For the next code sample I used a case class that extends Exception : case class JimOverslept(msg: String) extends Exception try { throw JimOverslept(msg = "went to bed late") } catch { case e: JimOverslept => println(e) } output: playground.CaseClassOutput$JimOverslept Actually, I

toString method for inherited case class in Scala

由 为君一笑 提交于 2020-05-16 06:32:06
问题 I am facing some inconsistency in calling toString method for case-classes in Scala. The first code sample: case class Person(name: String, age: Int) val jim = Person("jim", 42) println(jim) output: Person(jim,42) For the next code sample I used a case class that extends Exception : case class JimOverslept(msg: String) extends Exception try { throw JimOverslept(msg = "went to bed late") } catch { case e: JimOverslept => println(e) } output: playground.CaseClassOutput$JimOverslept Actually, I

java.lang.StackOverflowError when trying to add the same instance of list multiple times

由 爷，独闯天下 提交于 2020-04-12 07:42:31
问题 How to resolve java.lang.StackOverflowError for the following code? Person.java import java.util.List; public class Person { private String name; private List<Person> children; public Person() { } public Person(String name, List<Person> children) { this.name = name; this.children = children; } public String getName() { return name; } public void setName(String name) { this.name = name; } public List<Person> getChildren() { return children; } public void setChildren(List<Person> children) {

C#中string.Format()和ToString()格式化方法

由 泄露秘密 提交于 2020-04-08 07:26:23
C#数字格式化输出是我们在编程中经常需要处理的事情，那么这里向你介绍了一些C#数字格式化输出的例子，这样就会方便你来选择和比较，什么方式是比较适合自己项目的。 int a = 12345678; C#数字格式化之格式为sring输出 Label1.Text = string.Format("asdfadsf{0}adsfasdf",a); Label2.Text = "asdfadsf"+a.ToString()+"adsfasdf"; Label1.Text = string.Format("asdfadsf{0:C}adsfasdf",a);//asdfadsf￥1,234.00adsfasdf Label2.Text = "asdfadsf"+a.ToString("C")+"adsfasdf";//asdfadsf￥1,234.00adsfasdf double b = 1234.12543; int a = 12345678; C#数字格式化之格式为特殊的string样式输出 Label1.Text = string.Format("asdfadsf{0:C}adsfasdf",b);//asdfadsf￥1,234.13adsfasdf Label2.Text = "asdfadsf"+b.ToString("C")+"adsfasdf";//asdfadsf￥1
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