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graph-depends: Strip skeleton from dependency
skeleton being a mandatory dependency, we don't want all our packages to have a link back to that node, the graph would be awful. Signed-off-by: Maxime Hadjinlian <maxime.hadjinlian@gmail.com> Signed-off-by: Thomas Petazzoni <thomas.petazzoni@free-electrons.com>
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@ -271,16 +271,17 @@ def remove_transitive_deps(pkg,deps):
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new_d.append(d[i])
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new_d.append(d[i])
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return new_d
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return new_d
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# This function removes the dependency on the 'toolchain' package
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# This function removes the dependency on some 'mandatory' package, like the
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def remove_toolchain_deps(pkg,deps):
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# 'toolchain' package, or the 'skeleton' package
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return [p for p in deps[pkg] if not p == 'toolchain']
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def remove_mandatory_deps(pkg,deps):
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return [p for p in deps[pkg] if p not in ['toolchain', 'skeleton']]
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# This functions trims down the dependency list of all packages.
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# This functions trims down the dependency list of all packages.
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# It applies in sequence all the dependency-elimination methods.
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# It applies in sequence all the dependency-elimination methods.
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def remove_extra_deps(deps):
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def remove_extra_deps(deps):
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for pkg in list(deps.keys()):
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for pkg in list(deps.keys()):
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if not pkg == 'all':
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if not pkg == 'all':
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deps[pkg] = remove_toolchain_deps(pkg,deps)
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deps[pkg] = remove_mandatory_deps(pkg,deps)
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for pkg in list(deps.keys()):
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for pkg in list(deps.keys()):
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if not transitive or pkg == 'all':
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if not transitive or pkg == 'all':
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deps[pkg] = remove_transitive_deps(pkg,deps)
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deps[pkg] = remove_transitive_deps(pkg,deps)
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