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C++: fix used template function parameters

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It was missing colorizing and follow symbol template function parameters in case of:
* returning value
* use it as qualified name

Task-number: QTCREATORBUG-6861

Change-Id: I4226199e1f296cfe5a373783ebbc633e32fc9bcd
Reviewed-by: Erik Verbruggen <erik.verbruggen@digia.com>
This commit is contained in:
Przemyslaw Gorszkowski
2013-05-16 10:15:02 +02:00
committed by Erik Verbruggen
parent feff33827c
commit f93758b8e1
4 changed files with 49 additions and 2 deletions
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15
src/plugins/cpptools/cppchecksymbols.cpp
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@@ -408,6 +408,14 @@ Scope *CheckSymbols::enclosingScope() const
if (funDef->symbol)
return funDef->symbol;
} else if (TemplateDeclarationAST *templateDeclaration = ast->asTemplateDeclaration()) {
if (DeclarationAST *decl = templateDeclaration->declaration) {
if (FunctionDefinitionAST *funDef = decl->asFunctionDefinition()) {
if (funDef->symbol)
return funDef->symbol;
}
}
} else if (CompoundStatementAST *blockStmt = ast->asCompoundStatement()) {
if (blockStmt->symbol)
return blockStmt->symbol;
@@ -869,7 +877,12 @@ ClassOrNamespace *CheckSymbols::checkNestedName(QualifiedNameAST *ast)
const Name *name = class_or_namespace_name->name;
binding = _context.lookupType(name, enclosingScope());
addType(binding, class_or_namespace_name);
if (binding)
addType(binding, class_or_namespace_name);
else
// for the case when we use template parameter as qualifier
// e.g.: template <typename T> void fun() { T::type type; }
accept(nested_name_specifier->class_or_namespace_name);
for (it = it->next; it; it = it->next) {
NestedNameSpecifierAST *nested_name_specifier = it->value;