首页 \ 问答 \ 我可以通过复制它的内存克隆一个对象吗?(Can I clone an object by copying its memory?)

我可以通过复制它的内存克隆一个对象吗?(Can I clone an object by copying its memory?)

我需要在我的控制下为有限数量的类撤销+重做堆栈,这个类必须非常非常快,并且使用RTTI和XML或流是不可行的,因为嵌套对象中的实例数可能高达2000+名单。 对象需要通过备忘录模式复制进出,并立即重新加载。

有没有办法通过复制内存并重新实例化内存中的对象来克隆对象?


I need to have undo+redo stack for a limited number of classes under my control that has to be very very very fast and using RTTI and XML or streams is not feasible as the count of instances can be as high as 2000+ in nested object lists. The objects need to be copied into and out of via memento pattern and reloaded instantly.

Is there a way to clone objects by copying the memory and re-instantiating the objects from that memory?


原文:https://stackoverflow.com/questions/7798036
更新时间:2023-03-01 06:03

最满意答案

现在该程序打印Haha. Input sth here: Haha. Input sth here:并等待我的输入)。 如果我删除了scanf语句,它不在这里。 为什么?

由于标准( N1570 ..“几乎C11”)如此说,§5.1.2.3/ 6(强调我的):

对一致性实施的最低要求是:

[..]

  • 交互设备的输入和输出动态应按照7.21.3的规定进行。 这些要求的意图是尽可能快地出现无缓冲或线路缓冲输出,以确保提示消息实际上出现在等待输入的程序之前

[..]

即使您的输出不包含换行符,并且发送到缓冲的stdout行,它必须在您的程序被允许等待输入之前出现。 这是因为stdoutstdin连接到一个终端,因此( 注意:这是实现定义的! )标准称之为“交互式设备”。


Now the program prints Haha. Input sth here: (and wait for my input). It is not here if I remove the scanf statement. Why is that so?

Because the standard (N1570 .. "almost C11") says so, §5.1.2.3/6 (emphasis mine):

The least requirements on a conforming implementation are:

[..]

  • The input and output dynamics of interactive devices shall take place as specified in 7.21.3. The intent of these requirements is that unbuffered or line-buffered output appear as soon as possible, to ensure that prompting messages actually appear prior to a program waiting for input.

[..]

Even though your output does not contain a newline and is send to a line buffered stdout, it has to appear before your program is allowed to wait for input. This is because stdout and stdin are connected to a terminal and thus are (Attention: This is implementation defined!) what the standard calls "interactive devices".

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