C++ 在初始化时定义位集大小？
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Define bitset size at initialization?
提问by Martijn Courteaux
I want to make a bitset in C++. I did a bit of research. All examples I found where like this:
我想用 C++ 做一个 bitset。我做了一些研究。我发现的所有例子都是这样的：
bitset<6> myBitset;
// do something with it
But I don't know the size of the bitset when I define the variable in my class:
但是当我在类中定义变量时，我不知道 bitset 的大小：
#include <bitset>
class Test
{
public:
std::bitset *myBitset;
}
This won't compile...
这不会编译...
And initializing like this also doesn't work:
像这样初始化也不起作用：
int size = getDependentSizeForBitset();
myBitset = new bitset<size>();
采纳答案by jalf
Boost has a dynamic_bitsetyou can use.
Boost 有一个dynamic_bitset你可以使用的。
Alternatively, you can use a vector<bool>, which (unfortunately) is specialized to act as a bitset. This causes a lot of confusion, and in general is considered a bad idea. But that's how it works, so if that's what you need, you might as well use it, I suppose.
或者，您可以使用 a vector<bool>，它（不幸的是）专门用作位集。这会引起很多混乱，并且通常被认为是一个坏主意。但这就是它的工作原理，所以如果这是您的需要，我想您不妨使用它。
回答by Thomas Jones-Low
回答by graham.reeds
You should check out boosts dynamic_bitset.
您应该查看 boosts dynamic_bitset。
回答by AnT
What you are saying at the beginning is not true. The "examples you found" did not look as you posted. It is impossible to use a non-constant value to parametrize a template. So, your first example is invalid. Only constant expressions can serve as non-type arguments for a template. I.e. the non-type argument has to be a compile-time constant.
你一开始说的不是真的。“您找到的示例”看起来不像您发布的那样。不可能使用非常量值来参数化模板。所以，你的第一个例子是无效的。只有常量表达式可以作为模板的非类型参数。即非类型参数必须是编译时常量。
Of looks like you want to create a bitset whose size is not a compile-time constant. In this case the bitsettemplate is out of question. You need an implementation of run-time sized bitset. For example, you can use std::vector<bool>- in many (if not all) implementations this template is specialized to implement a packedarray of boolean values, where each element occupies one bit (as opposed to an boolobject).
看起来您想创建一个大小不是编译时常量的位集。在这种情况下，bitset模板是不可能的。您需要一个运行时大小的位集的实现。例如，您可以使用std::vector<bool>- 在许多（如果不是全部）实现中，此模板专门用于实现布尔值的打包数组，其中每个元素占用一位（与bool对象相反）。
回答by stinky472
bitset requires size as a template parameter, meaning the size has to be capable of being determined at compile-time. It cannot be based on a runtime condition, like user input.
bitset 需要 size 作为模板参数，这意味着必须能够在编译时确定大小。它不能基于运行时条件，例如用户输入。
For that, you should look into std::vector or boost::dynamic_bitset. std::vector is a specialized template instantiation that uses one bit per element. Unlike bitset, it can be dynamically sized.
为此，您应该查看 std::vector 或 boost::dynamic_bitset。std::vector 是一个专门的模板实例化，每个元素使用一位。与 bitset 不同，它可以动态调整大小。
回答by stinky472
You can make your class a template to make the std::bitsetsize undetermined until your class gets instantiated. You can do it like this:
你可以让你的类成为一个模板，在std::bitset你的类被实例化之前不确定大小。你可以这样做：
#include <bitset>
template<int size>
class Test
{
public:
std::bitset<size> bitset;
//...
}
Then to use your class you would have to do this:
然后要使用您的课程，您必须这样做：
int exampleSize = 42;
Test<exampleSize> name;
