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41 lines
1.9 KiB
ReStructuredText
41 lines
1.9 KiB
ReStructuredText
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Copy policy in building expressions [Advanced]
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When building expressions, a problem can appear for the object at the leaves of the expressions tree
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(scalar, placeholders, various callable objects, etc...).
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The question is whether they should be captured by value or by reference.
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In the lazy library, the choice has been made to capture them **by value**.
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This means that, by default, *all objects appearing in a lazy expression are copied, rather than captured by reference*.
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This is necessary to store expressions (with auto like above) for future reuse, transform them into new expressions
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(e.g. make partial evaluation). Expressions are objects.
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Capturing the leaves by reference easily produces dangling references.
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[Basically, an expression is a function of its placeholder. The leaves are the parameters of the function,
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they must leave as long as the expression live. In other words, we need to have a closure of the function].
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The drawback of this approach is that it can generate unless copies of large objects.
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There are several solutions to this issue :
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* The object is very small (like a double), hence making a copy in not a problem.
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* You *know* that the object `A` will survive the expression, so using a reference is not a problem.
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You can use the `lazy(A)` expression that will wrap the reference to the object.
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* The object has a compagnon view object (like array, array_view). In this case,
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one wishes to put a view of the object rather than a copy in the expression.
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There are two cases.
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* TRIQS objects like array, matrix, vector will do it automatically.
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* For another object, if the object defines the tag `has_view_type_tag` as ::
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typedef void has_view_type_tag;
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typedef MY_VIEW_TYPE view_type;
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the instance of the object will be replaced by an instance of its view_type in building the expression.
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For an illustration, Cf....
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