“回滚”或撤消任何应用于流的操纵器，而无需知道操纵器是什么 [英] "Roll-Back" or Undo Any Manipulators Applied To A Stream Without Knowing What The Manipulators Were

问题描述

如果我对流应用任意数量的操纵器，是否有办法以通用方式撤消那些操纵器的应用？

If I apply an arbitrary number of manipulators to a stream, is there a way to undo the application of those manipulators in a generic way?

例如，考虑以下内容：

#include <iostream>
#include <iomanip>

using namespace std;

int main()
{
    cout << "Hello" << hex << 42 << "\n";
    // now i want to "roll-back" cout to whatever state it was in
    // before the code above, *without* having to know 
    // what modifiers I added to it

    // ... MAGIC HAPPENS! ...

    cout << "This should not be in hex: " << 42 << "\n";
}

假设我想在 MAGIC HAPPENS 会将流操纵器的状态恢复到我做 cout<<之前的状态。十六进制。 但是我不知道我添加了哪些操纵器。我怎么能做到这一点？

Suppose I want to add code at MAGIC HAPPENS that will revert the state of the stream manipulators to whatever it was before I did cout << hex. But I don't know what manipulators I added. How can I accomplish this?

换句话说，我希望能够写出这样的东西（伪代码/幻想代码）：

In other words, I'd like to be able to write something like this (psudocode/fantasy code):

std::something old_state = cout.current_manip_state();
cout << hex;
cout.restore_manip_state(old_state);

这可能吗？

如果您很好奇，我有兴趣在自定义的 operator<<（）我正在写一个复杂的类型。该类型是一种有区别的联合，并且不同的值类型将对流应用不同的操作。

In case you're curious, I'm interested in doing this in a custom operator<<() I'm writing for a complex type. The type is a kind of discriminated union, and different value types will have different manips applied to the stream.

限制：我不能使用Boost或任何其他第三方库。解决方案必须使用标准C ++。

Restriction: I cannot use Boost or any other 3rd party libraries. Solution must be in standard C++.

推荐答案

是。

您可以保存状态并恢复状态：

You can save the state and restore it:

#include <iostream>
#include <iomanip>

using namespace std;

int main()
{

    std::ios  state(NULL);
    state.copyfmt(std::cout);

    cout << "Hello" << hex << 42 << "\n";
    // now i want to "roll-back" cout to whatever state it was in
    // before the code above, *without* having to know what modifiers I added to it

  // ... MAGIC HAPPENS! ...

    std::cout.copyfmt(state);
    cout << "This should not be in hex: " << 42 << "\n";
}

如果您想返回默认状态，则甚至不需要要保存状态，可以从临时对象中提取状态。

If you want to get back to the default state you don't even need to save the state you can extract it from a temporary object.

std::cout.copyfmt(std::ios(NULL));

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