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Merge bitcoin#28452: Do not use std::vector = {} to release memory
3fcd7fc Do not use std::vector = {} to release memory (Pieter Wuille)
Pull request description:
It appears that invoking `v = {};` for an `std::vector<...> v` is equivalent to `v.clear()`, which does not release its allocated memory. There are a number of places in the codebase where it appears to be used for that purpose however (mostly written by me). Replace those with `std::vector<...>{}.swap(v);` (using a helper function `ClearShrink` in util/vector.h).
To explain what is going on: `v = {...};` is equivalent in general to `v.operator=({...});`. For many types, the `{}` is converted to the type of `v`, and then assigned to `v` - which for `std::vector` would ordinarily have the effect of clearing its memory (constructing a new empty vector, and then move-assigning it to `v`). However, since `std::vector<T>` has an `operator=(std::initializer_list<T>)` defined, it has precedence (since no implicit conversion is needed), and with an empty list, that is equivalent to `clear()`.
I did consider using `v = std::vector<T>{};` as replacement for `v = {};` instances where memory releasing is desired, but it appears that it does not actually work universally either. `V{}.swap(v);` does.
ACKs for top commit:
ajtowns:
utACK 3fcd7fc
stickies-v:
ACK 3fcd7fc
theStack:
Code-review ACK 3fcd7fc
Tree-SHA512: 6148558126ec3c8cfd6daee167ec1c67b360cf1dff2cbc132bd71768337cf9bc4dda3e5a9cf7da4f7457d2123288eeba77dd78f3a17fa2cfd9c6758262950cc5
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