C++ 部分模板专业化和 Natvis
[英]C++ Partial Template Specialization and Natvis
问题描述
I'm trying to create Visual Studio debug visualizers for a partially specialized type.
我正在尝试为部分专业化类型创建 Visual Studio 调试可视化工具。 For example, let's say I have something like this:例如，假设我有这样的事情：
template <typename T>
struct Foo
{
T bar;
};
template <typename T>
struct Foo<T*>
{
T baz;
};
Without the partial specialization, this would be easy:如果没有部分专业化，这将很容易：
<Type Name="Foo<*>"> ... </Type>
With full specialization, it'd also be easy:有了完全的专业化，这也很容易：
<Type Name="Foo<int>"> ... </Type>
But how do I cover the partial specialization?但是我如何涵盖部分专业化？ Is this even supported?这甚至被支持吗？ If not, is there a workaround?如果没有，是否有解决方法？
1 个解决方案
解决方案1
2 已采纳 2019-11-07 07:35:06
Short answer - No. You can't specify type qualifiers, references, and so on in natvis type name <Type Name="Foo<*>"> .简短回答 - 不。您不能在 natvis 类型名称<Type Name="Foo<*>">中指定类型限定符、引用等。
But:但：
You can use the template's typename parameter as a string and compare with a type.您可以将模板的 typename 参数用作字符串并与类型进行比较。 For example, in the node's Condition attribute:例如，在节点的Condition属性中：
<Type Name="Foo<*>">
<DisplayString Condition="strcmp("$T1","short")==0">specialization short</DisplayString>
<DisplayString Condition="strcmp("$T1","int &")==0">specialization int &</DisplayString>
<DisplayString>unspecified specialization</DisplayString>
</Type>
For Foo<short> you will see specialization short and unspecified specialization for other.对于Foo<short> ，您将看到其他的specialization short和unspecified specialization 。
Example:例子：
template <typename T, typename U>
struct Foo
{
T bar;
};
template <typename U>
struct Foo<int &, U>
{
U baz;
};
int main()
{
int gg = 0;
Foo<short, int> a;
Foo<int, int> b;
Foo<int &, int> c;
Natvis:纳特维斯：
<Type Name="Foo<*,*>" >
<DisplayString Condition="strcmp("$T1","short")==0">specialization short</DisplayString>
<DisplayString>unspecified specialization</DisplayString>
</Type>
<Type Name="Foo<int&,*>">
<DisplayString>partial specialization int&</DisplayString>
</Type>
Result:结果：
Or you, if you have some unique member in your partially specialized type, can use Priority option.或者，如果您的部分专业类型中有一些独特的成员，则可以使用Priority选项。
Example:例子：
template <typename T>
struct Foo
{
T bar;
};
template <typename U>
struct Foo<U &>
{
U baz;
};
int main()
{
int g = 0;
Foo<short> a;
Foo<int> b;
Foo<int &> c{g};
Natvis:纳特维斯：
<Type Name="Foo<*>">
<DisplayString>partial specialization {baz}</DisplayString>
</Type>
<Type Name="Foo<*>" Priority="Low">
<DisplayString>other specialization</DisplayString>
</Type>
Result:结果：
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