本文整理汇总了C++中PythonObject::ResolveName方法的典型用法代码示例。如果您正苦于以下问题：C++ PythonObject::ResolveName方法的具体用法？C++ PythonObject::ResolveName怎么用？C++ PythonObject::ResolveName使用的例子？那么, 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PythonObject的用法示例。
在下文中一共展示了PythonObject::ResolveName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: ResolveName
PythonObject PythonObject::ResolveName(llvm::StringRef name) const {
// Resolve the name in the context of the specified object. If, for example,
// `this` refers to a PyModule, then this will look for `name` in this
// module. If `this` refers to a PyType, then it will resolve `name` as an
// attribute of that type. If `this` refers to an instance of an object,
// then it will resolve `name` as the value of the specified field.
//
// This function handles dotted names so that, for example, if `m_py_obj`
// refers to the `sys` module, and `name` == "path.append", then it will find
// the function `sys.path.append`.
size_t dot_pos = name.find('.');
if (dot_pos == llvm::StringRef::npos) {
// No dots in the name, we should be able to find the value immediately as
// an attribute of `m_py_obj`.
return GetAttributeValue(name);
}
// Look up the first piece of the name, and resolve the rest as a child of
// that.
PythonObject parent = ResolveName(name.substr(0, dot_pos));
if (!parent.IsAllocated())
return PythonObject();
// Tail recursion.. should be optimized by the compiler
return parent.ResolveName(name.substr(dot_pos + 1));
}示例2:
PythonObject
PythonObject::ResolveNameWithDictionary(llvm::StringRef name,
const PythonDictionary &dict) {
size_t dot_pos = name.find('.');
llvm::StringRef piece = name.substr(0, dot_pos);
PythonObject result = dict.GetItemForKey(PythonString(piece));
if (dot_pos == llvm::StringRef::npos) {
// There was no dot, we're done.
return result;
}
// There was a dot. The remaining portion of the name should be looked up in
// the context of the object that was found in the dictionary.
return result.ResolveName(name.substr(dot_pos + 1));
}