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https://github.com/bazelbuild/bazel-skylib
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1b28145983
In this PR: - In the _resolve_locations function: use the Bash-less ctx.resolve_tools function to resolve the runfiles manifests and inputs of tools, instead of using ctx.resolve_command for the same purpose. - In the _custom_envmap function: no longer resolve $(location) references when creating the envvars from custom_env, because those references were already resolved in _resolve_locations. The ctx.resolve_tools() method was added in this PR: https://github.com/bazelbuild/bazel/pull/7139 See design doc there.
69 lines
2.7 KiB
Python
69 lines
2.7 KiB
Python
# Copyright 2019 The Bazel Authors. All rights reserved.
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#
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# Licensed under the Apache License, Version 2.0 (the "License");
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# you may not use this file except in compliance with the License.
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# You may obtain a copy of the License at
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#
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# http://www.apache.org/licenses/LICENSE-2.0
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#
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# Unless required by applicable law or agreed to in writing, software
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# distributed under the License is distributed on an "AS IS" BASIS,
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# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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# See the License for the specific language governing permissions and
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# limitations under the License.
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"""Utilities for maprule."""
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def resolve_locations(ctx, strategy, d):
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"""Resolve $(location) references in the values of a dictionary.
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Args:
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ctx: the 'ctx' argument of the rule implementation function
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strategy: a struct with an 'as_path(string) -> string' function
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d: {string: string} dictionary; values may contain $(location) references
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for labels declared in the rule's 'srcs' and 'tools' attributes
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Returns:
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{string: string} dict, same as 'd' except "$(location)" references are
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resolved.
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"""
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location_expressions = []
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parts = {}
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was_anything_to_resolve = False
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for k, v in d.items():
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# Look for "$(location ...)" or "$(locations ...)", resolve if found.
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# _validate_attributes already ensured that there's at most one $(location/s ...) in "v".
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if "$(location" in v:
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tokens = v.split("$(location")
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was_anything_to_resolve = True
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closing_paren = tokens[1].find(")")
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location_expressions.append("$(location" + tokens[1][:closing_paren + 1])
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parts[k] = (tokens[0], tokens[1][closing_paren + 1:])
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else:
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location_expressions.append("")
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resolved = {}
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if was_anything_to_resolve:
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# Resolve all $(location) expressions in one go. Should be faster than resolving them
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# one-by-one.
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all_location_expressions = "<split_here>".join(location_expressions)
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all_resolved_locations = ctx.expand_location(all_location_expressions)
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resolved_locations = strategy.as_path(all_resolved_locations).split("<split_here>")
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i = 0
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# Starlark dictionaries have a deterministic order of iteration, so the element order in
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# "resolved_locations" matches the order in "location_expressions", i.e. the previous
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# iteration order of "d".
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for k, v in d.items():
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if location_expressions[i]:
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head, tail = parts[k]
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resolved[k] = head + resolved_locations[i] + tail
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else:
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resolved[k] = v
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i += 1
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else:
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resolved = d
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return resolved
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