Why would one replace default new and delete operators?

Why should would one replace the default operator new and delete with a custom new and delete operators? This is in continuation of Overloading new and delete in the immensely illuminating C++ FAQ: Operator overloading. An followup entry to this FAQ is: How should I write ISO C++ standard conformant custom new and delete operators? Note: The answer is based on lessons from Scott Meyers&

为什么要替换默认的新运算符和删除运算符?

为什么人会替换默认的运营商new和delete一个自定义new和delete运营商? 这是在极其明亮的C ++ FAQ中重载新的和删除的继续: 运算符超载。 此常见问题解答的后续条目是: 我应该如何编写符合ISO C ++标准的自定义new和delete操作符? 注意:答案是基于Scott Meyers的More Effective C ++的经验教训。 (注意:这是一个Stack Overflow的C ++常见问题解答的入口,如果你想批评在这个表单中提供常见问题的想法,那么开始

What are rvalues, lvalues, xvalues, glvalues, and prvalues?

In C++03, an expression is either an rvalue or an lvalue . In C++11, an expression can be an: rvalue lvalue xvalue glvalue prvalue Two categories have become five categories. What are these new categories of expressions? How do these new categories relate to the existing rvalue and lvalue categories? Are the rvalue and lvalue categories in C++0x the same as they are in C++03?

什么是右值,左值,xvalues,glvalues和prvalues?

在C ++ 03中,表达式是右值或左值 。 在C ++ 11中,表达式可以是: 右值 左值 x值 glvalue prvalue 两类分为五类。 这些新的表达类别是什么? 这些新类别如何与现有的右值和左值类别相关联? C ++ 0x中的右值和左值类别与C ++ 03中的相同吗? 为什么需要这些新类别? WG21神只是试图将我们这些凡人混为一谈? 我想这个文件可能不是那么简短的介绍:n3055 整个屠杀始于移动语义。 一旦我们有了可以移

What is the copy

What is this idiom and when should it be used? Which problems does it solve? Does the idiom change when C++11 is used? Although it's been mentioned in many places, we didn't have any singular "what is it" question and answer, so here it is. Here is a partial list of places where it was previously mentioned: What are your favorite C++ Coding Style idioms: Copy-swap Copy

什么是副本

什么是这个成语,什么时候应该使用它? 它解决了哪些问题? 当使用C ++ 11时,习语是否会发生变化? 尽管在很多地方都有提到,但我们没有任何单数的“问题”和答案,所以在这里。 这是以前提到的地方的部分列表: 什么是您最喜爱的C ++编码风格习语:复制交换 在C ++中复制构造函数和=运算符重载:是一个常用函数吗? 什么是复制省略以及它如何优化复制 - 交换习惯用法 C ++:动态分配一个对象数组? 概观 为什么

What is the difference between these two for loops in C?

This question already has an answer here: What is the “-->” operator in C++? 21 answers As for the first one, i is decremented before the loop body is executed. The second one decrements i after the loop body is executed. The difference is the step in which i is actually decremented, which affects the values of i seen inside the loop body. The second traditional version decrements i

这两个for循环在C中有什么区别?

这个问题在这里已经有了答案: 什么是C ++中的“ - >”运算符? 21个答案 至于第一个, i在执行循环体之前递减。 第二个在循环体执行后递减i 。 不同的是i实际上递减的步骤,它会影响i在循环体内看到的值。 第二个传统版本在循环体执行后递减i ,并且在条件再次被检查之前。 因此,在i == 1执行循环体后, i达到了0。 条件再次检查并且在循环i为0之后。 作为检查条件的一部分,第一个版本在循环体执行之前递减i

Meaning of "+ +" operator (not ++)

This question already has an answer here: Several unary operators in C and C++ 3 answers What is the “-->” operator in C++? 21 answers Unary plus (+) against literal string 1 answer There is no + + operator. There's a + operator (which occurs in both unary and binary forms), and a ++ operator, not used here. Each of those is a binary + operator followed by one or more unary + o

“+ +”运算符的含义(不是++)

这个问题在这里已经有了答案: C和C ++ 3中有几个一元运算符 什么是C ++中的“ - >”运算符? 21个答案 一元加(+)针对文字串1的答案 没有+ +运算符。 有一个+运算符(以一元和二元形式出现)和一个++运算符,这里不使用。 每个那些的是二进制+运算符,随后通过一个或多个一元+运算符。 这个: c = c + + "d"; 相当于 c = c + (+ "d"); 这个: c = c + + + "d"; 相当于: c = c + (+ + "d"); 要么: c =

>> actually do?

This question already has an answer here: What is the “-->” operator in C++? 21 answers while( x -->> 0 ) // x runs to 0 This is actually a hybrid of the -- (post-decrement) and >> (bitshift right) operators, better formatted as: while (x-- >> 0) For this specific usage, with 0 on the right hand side, x is decremented with each loop iteration due to the postfix -- ,

>>其实呢?

这个问题在这里已经有了答案: 什么是C ++中的“ - >”运算符? 21个答案 while(x - >> 0)// x运行到0 这实际上是-- (后递减)和>> (右移位)运算符的混合体,格式更好: while (x-- >> 0) 对于这个特定的用法,在右边为0时,由于后缀-- ,每次循环迭代x递减,并且先前(递减前)值被>> 0右移0位,其根本不执行任何操作当x不是负数时。 当x是1时,后减量将其减小到0,并且移位的结果

Profiling help required

I have a profiling issue - imagine I have the following code... void main() { well_written_function(); badly_written_function(); } void well_written_function() { for (a small number) { highly_optimised_subroutine(); } } void badly_written_function() { for (a wastefully and unnecessarily large number) { highly_optimised_subroutine(); } } void highly

需要分析帮助

我有一个分析问题 - 想象我有下面的代码... void main() { well_written_function(); badly_written_function(); } void well_written_function() { for (a small number) { highly_optimised_subroutine(); } } void badly_written_function() { for (a wastefully and unnecessarily large number) { highly_optimised_subroutine(); } } void highly_optimised_subroutine()

Why should C++ programmers minimize use of 'new'?

I stumbled upon Stack Overflow question Memory leak with std::string when using std::list<std::string>, and one of the comments says this: Stop using new so much. I can't see any reason you used new anywhere you did. You can create objects by value in C++ and it's one of the huge advantages to using the language. You do not have to allocate everything on the heap. Stop thinkin

为什么C ++程序员应该尽量减少对'新'的使用?

我偶然发现堆栈溢出问题使用std :: list <std :: string>时使用std :: string的内存泄漏,其中一个注释说: 停止使用new东西。 我看不出任何你在任何地方使用新的理由。 您可以在C ++中按值创建对象,这是使用该语言的巨大优势之一。 您不必分配堆中的所有内容。 不要像Java程序员那样思考。 我不确定他的意思。 为什么应该尽可能经常通过C ++的价值创建对象,它在内部会产生什么差异? 我误解了答案吗? 有两

What is the effect of extern "C" in C++?

What exactly does putting extern "C" into C++ code do? For example: extern "C" { void foo(); } extern "C" makes a function-name in C++ have 'C' linkage (compiler does not mangle the name) so that client C code can link to (ie use) your function using a 'C' compatible header file that contains just the declaration of your function. Your function definiti

C ++中extern“C”的效果是什么?

把extern "C"放到C ++代码中究竟做了什么? 例如: extern "C" { void foo(); } extern“C”使C ++中的函数名具有'C'链接(编译器不会破坏名称),以便客户端C代码可以使用'C'兼容头文件链接到(即使用)您的函数,该文件仅包含你的函数声明。 您的函数定义包含在二进制格式中(由C ++编译器编译),客户端的“C”链接器将使用“C”名称链接到该格式。 由于C ++重载了函数名,而C没有,C ++编译器

Meaning of "const" last in a C++ method declaration?

What is the meaning of const in declarations like these? The const confuses me. class foobar { public: operator int () const; const char* foo() const; }; When you add the const keyword to a method the this pointer will essentially become a pointer to const object, and you cannot therefore change any member data. (Unless you use mutable , more on that later). The const keyword is

C ++方法声明中最后一个“const”的含义?

const在这些声明中的含义是什么? const让我困惑。 class foobar { public: operator int () const; const char* foo() const; }; 将const关键字添加到方法时, this指针本质上将成为指向const对象的指针,因此不能更改任何成员数据。 (除非你使用mutable ,稍后更多)。 const关键字是函数签名的一部分,这意味着您可以实现两个类似的方法,一个是在对象为const时调用的,另一个不是。 #include <iostrea