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C++: fix scope matching for templates in FindUsages

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When the cursor is on the name of declaration of a templated function,
then since f93758b8e1 the scope returned
by Document::findScopeAt is the scope of the template declaration.
Before it was the parent scope of the template declaration.

The check in FindUsages::checkCandidates did not check all combinations
of template(-child symbol) scopes for the searched symbol and its
occurrences.

Task-number: QTCREATORBUG-9749

Change-Id: Idc84a2ba718721ce54683a67635a93352784ddd1
Reviewed-by: Nikolai Kosjar <nikolai.kosjar@digia.com>
This commit is contained in:
Erik Verbruggen
2013-07-08 12:30:06 +02:00
committed by Nikolai Kosjar
parent 07c0a8348c
commit d70a33c0d0
2 changed files with 40 additions and 2 deletions
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10
src/libs/cplusplus/FindUsages.cpp
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@@ -208,7 +208,15 @@ bool FindUsages::checkCandidates(const QList<LookupItem> &candidates) const
if (isLocalScope(_declSymbol->enclosingScope()) || isLocalScope(s->enclosingScope())) {
if (s->enclosingScope()->isTemplate()) {
if (s->enclosingScope()->enclosingScope() != _declSymbol->enclosingScope())
if (_declSymbol->enclosingScope()->isTemplate()) {
if (s->enclosingScope()->enclosingScope() != _declSymbol->enclosingScope()->enclosingScope())
return false;
} else {
if (s->enclosingScope()->enclosingScope() != _declSymbol->enclosingScope())
return false;
}
} else if (_declSymbol->enclosingScope()->isTemplate() && s->isTemplate()) {
if (_declSymbol->enclosingScope()->enclosingScope() != s->enclosingScope())
return false;
} else if (! s->isUsingDeclaration() && s->enclosingScope() != _declSymbol->enclosingScope()) {
return false;