本文整理汇总了C++中TypePtr::couldBe方法的典型用法代码示例。如果您正苦于以下问题：C++ TypePtr::couldBe方法的具体用法？C++ TypePtr::couldBe怎么用？C++ TypePtr::couldBe使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类TypePtr的用法示例。
在下文中一共展示了TypePtr::couldBe方法的2个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: Inferred
/* We have inferred type1 and type2 as the actual types for the same
expression.
Check that the types are compatible (it cant be both a string and
an integer, for example), and return the combined type. If they
are not compatible, return a null pointer.
*/
TypePtr Type::Inferred(AnalysisResultConstPtr ar,
TypePtr type1, TypePtr type2) {
if (!type1) return type2;
if (!type2) return type1;
KindOf k1 = type1->m_kindOf;
KindOf k2 = type2->m_kindOf;
if (k1 == k2) {
return k1 == KindOfObject ?
Type::InferredObject(ar, type1, type2) : type1;
}
// If one set is a subset of the other, return the subset.
if ((k1 & k2) == k1) return type1;
if ((k1 & k2) == k2) return type2;
// If one type must be numeric and the other might be, then assume numeric
if (type1->mustBe(KindOfNumeric) && type2->couldBe(KindOfNumeric)) {
return type1;
}
if (type2->mustBe(KindOfNumeric) && type1->couldBe(KindOfNumeric)) {
return type2;
}
// Otherwise, take the intersection
int resultKind = type1->m_kindOf & type2->m_kindOf;
if (resultKind == KindOfObject) {
return Type::InferredObject(ar, type1, type2);
}
return resultKind ? GetType(resultKind) : TypePtr();
}示例2: Inferred
/* We have inferred type1 and type2 as the actual types for the same
expression.
Check that the types are compatible (it cant be both a string and
an integer, for example), and return the combined type. If they
are not compatible, return a null pointer.
*/
TypePtr Type::Inferred(AnalysisResultConstPtr ar,
TypePtr type1, TypePtr type2) {
if (!type1) return type2;
if (!type2) return type1;
KindOf k1 = type1->m_kindOf;
KindOf k2 = type2->m_kindOf;
if (k1 == k2) return type1;
// If one set is a subset of the other, return the subset.
if ((k1 & k2) == k1) return type1;
if ((k1 & k2) == k2) return type2;
// If one type must be numeric and the other might be, then assume numeric
if (type1->mustBe(KindOfNumeric) && type2->couldBe(KindOfNumeric))
return type1;
if (type2->mustBe(KindOfNumeric) && type1->couldBe(KindOfNumeric))
return type2;
// Otherwise, take the intersection
int resultKind = type1->m_kindOf & type2->m_kindOf;
std::string resultName = "";
if (resultKind & KindOfObject) {
// if they're the same, or we don't know one's name, then use
// the other
if (type1->m_name == type2->m_name || type1->m_name.empty()) {
resultName = type2->m_name;
} else if (type2->m_name.empty()) {
resultName = type1->m_name;
} else {
// take the subclass
ClassScopePtr cls1 = ar->findClass(type1->m_name);
ClassScopePtr cls2 = ar->findClass(type2->m_name);
if (cls1 && !cls1->isRedeclaring()
&& cls1->derivesFrom(ar, type2->m_name, true, false)) {
resultName = type1->m_name;
} else if (cls2 && !cls2->isRedeclaring()
&& cls2->derivesFrom(ar, type1->m_name, true, false)) {
resultName = type2->m_name;
} else {
resultKind &= ~KindOfObject;
}
}
}
if (resultKind)
return TypePtr(new Type((KindOf)resultKind, resultName));
else
return TypePtr();
}