fix: resolve namespace while instantiating templates in a namespace #151
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clingwrapper/src/clingwrapper.cxx
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| @@ -593,6 +593,17 @@ bool Cppyy::AppendTypesSlow(const std::string& name, | |||
| if (is_integral(i)) | |||
| integral_value = strdup(i.c_str()); | |||
| types.emplace_back(type, integral_value); | |||
| } else if (parent && (Cpp::IsNamespace(parent) || Cpp::IsClass(parent))) { | |||
| std::string qualified_name = Cpp::GetQualifiedCompleteName(parent) + "::" + name; | |||
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Can we replace that with a simple lookup without qualification.
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Nope, the lookup does not work without the qualification.
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We need to figure out what api we need to create to avoid string concatenation.
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I believe this is acceptable. We are only concatenating the parent scope with the child. There is nothing else going on here.
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Why GetNamed(name, parent) does not work?
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GetNamed worked. I initially tried GetScope and it failed. I have pushed the changes.
Vipul-Cariappa
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clingwrapper/src/clingwrapper.cxx
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| @@ -607,22 +617,16 @@ Cppyy::TCppType_t Cppyy::GetType(const std::string &name, bool enable_slow_looku | |||
| if (auto type = Cpp::GetType(name)) | |||
| return type; | |||
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| if (!enable_slow_lookup) { | |||
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I think that diagnostic is important to expose interfaces that still do parsing of qualified names.
Enables instantiating templates `f` with type `t` without explicitly specifying the namespace in type `t`, if `f` and `t` belong to the same namespace.
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Enables instantiating templates
fwith typetwithout explicitly specifying the namespace in typet, iffandtbelong to the same namespace.Has a cyclic dependency with compiler-research/CPyCppyy#104