compound-assignment

Why doesn't Java have compound assignment versions of the conditional-and and conditional-or operators? (&&=, ||=)

由 試著忘記壹切 提交于 2019-12-17 03:03:42
问题 So for binary operators on booleans, Java has & , | , ^ , && and || . Let's summarize what they do briefly here: JLS 15.22.2 Boolean Logical Operators &, ^, and | JLS 15.23 Conditional-And Operator && JLS 15.24 Conditional-Or Operator || For & , the result value is true if both operand values are true ; otherwise, the result is false . For | , the result value is false if both operand values are false ; otherwise, the result is true . For ^ , the result value is true if the operand values are

What exactly does += do in python?

由 谁都会走 提交于 2019-12-16 22:35:30
问题 I need to know what += does in python. It's that simple. I also would appreciate links to definitions of other short hand tools in python. 回答1: In Python, += is sugar coating for the __iadd__ special method, or __add__ or __radd__ if __iadd__ isn't present. The __iadd__ method of a class can do anything it wants. The list object implements it and uses it to iterate over an iterable object appending each element to itself in the same way that the list's extend method does. Here's a simple

Compound assignment and add operator overloading

由 我怕爱的太早我们不能终老 提交于 2019-12-11 17:15:11
问题 I need help with both of my operator overloading functions presented below. I'm unsure of how I can implement this without actually using the assignment in the function definitions. Code for operator + in my .cpp file: MyString& MyString::operator +(const MyString& rhs) { delete [] String; String = new char[rhs.Size]; Size = rhs.Size; // needs to be a loop for cascading + // so that String1=String2+String3+String4 will work for(int i = 0; i < rhs.Size+1 ; i++) { // String[i] + rhs.String[i];

Java Compound Assignment Operator precedence in expressions

由 守給你的承諾、 提交于 2019-12-11 04:56:43
问题 The output of the following code is declared as "6" when I try to execute this. When I am trying to think through this, the expression "k += 3 + ++k; should have been evaluated as k = k + (3 + ++k); but in this case the output should have been 7. Looks like it was evaluated as k = k + 3 + ++k; which resulted in 6. Could someone please explain me why the expression was evaluated as "k + 3 + ++k" instead of " k + (3 + ++k); ? public class TestClass { public static int m1(int i){ return ++i; }

Are compound statements lvalue (or rvalue) in C?

由 匆匆过客 提交于 2019-12-10 17:46:35
问题 When I examined the definition of the container_of macro in Linux Kernel, I saw a compound statement as a macro definition, #define container_of(ptr, type, member) ({ \ const typeof( ((type *)0)->member ) *__mptr = (ptr); \ (type *)( (char *)__mptr - offsetof(type,member) );}) however the unclear issue in my mind is which statement gets taken into account as rvalue. Apparently result of the last statement is used as rvalue, but why? (type *)( (char *)__mptr - offsetof(type,member) ); For

Assigning using += gives NaN in javascript

由 拥有回忆 提交于 2019-12-10 10:06:13
问题 Assignment a number to an attribute using the += operator gives me NaN in JavaScript. This code works as expected: > var result = {}; undefined > result['value'] = 10; 10 > result['value'] += 10; 20 But here we get NaN : > var test = {}; undefined > test['value'] += 10; NaN Why does JavaScript behave like this? How can I get this to work without initializing result['value'] = 0 ? 回答1: You can't add a number to undefined in JavaScript. If you don't want to initialize the number, you need to

Auto-(un)boxing fail for compound assignment

由 让人想犯罪 __ 提交于 2019-12-08 16:37:19
问题 Thanks to the implicit casting in compound assignments and increment/decrement operators, the following compiles: byte b = 0; ++b; b++; --b; b--; b += b -= b *= b /= b %= b; b <<= b >>= b >>>= b; b |= b &= b ^= b; And thanks to auto-boxing and auto-unboxing, the following also compiles: Integer ii = 0; ++ii; ii++; --ii; ii--; ii += ii -= ii *= ii /= ii %= ii; ii <<= ii >>= ii >>>= ii; ii |= ii &= ii ^= ii; And yet, the last line in the following snippet gives compile-time error: Byte bb = 0;

Assigning using += gives NaN in javascript

由 六月ゝ 毕业季﹏ 提交于 2019-12-06 01:22:24
Assignment a number to an attribute using the += operator gives me NaN in JavaScript. This code works as expected: > var result = {}; undefined > result['value'] = 10; 10 > result['value'] += 10; 20 But here we get NaN : > var test = {}; undefined > test['value'] += 10; NaN Why does JavaScript behave like this? How can I get this to work without initializing result['value'] = 0 ? Blazemonger You can't add a number to undefined in JavaScript. If you don't want to initialize the number, you need to test if it's undefined before incrementing it: test['value'] = (typeof test['value']==='undefined'

Difference between string += s1 and string = string + s1 [closed]

由 纵然是瞬间 提交于 2019-12-03 10:39:09
问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 4 months ago . One of my programs is exceeding the time limit when I am using fans = fans + s[i] , while when I am using fans += s[i] it is being accepted... Why does this happen? To Explain more , fans is a string and s is also a string so while iterating over string s i want only some characters of s so i am creating a new

Difference between string += s1 and string = string + s1 [closed]

由 喜夏-厌秋 提交于 2019-12-03 01:11:07
One of my programs is exceeding the time limit when I am using fans = fans + s[i] , while when I am using fans += s[i] it is being accepted... Why does this happen? To Explain more , fans is a string and s is also a string so while iterating over string s i want only some characters of s so i am creating a new string fans.Now there are two ways in which i can add character to my new string fans. The Problem is mentioned below fans = fans + s[i]; // gives Time limit exceeded fans += s[i]; // runs successfully For built-in types a += b is exactly the same as a = a + b , but for classes, those
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