docs: fix arithmetic update-assignment equivalence (#3682)
The manual said `a op= b` is equivalent to `a |= . op b`. It is actually equivalent to `(b as $b | a |= . op $b)`: the right-hand side gets `.` as input, not the value being updated, and each value it outputs gives a separate result. This updates the description and adds two examples, `.a += .b` and `.a += (1,2)`. `tests/man.test` and `jq.1.prebuilt` are regenerated. Fixes #2119, fixes #2407
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@@ -3561,6 +3561,13 @@ sections:
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The left-hand side can be any general path expression; see `path()`.
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The right-hand side of `|=` is run with the value at the path
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as its input, not with `.`. This is what sets it apart from
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plain assignment (`=`) and the arithmetic update-assignment
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operators such as `+=`, whose right-hand side is run with `.`.
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For example, `.a |= .b` sets `.a` to the `b` field of `.a`,
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while `.a = .b` and `.a += .b` use the `b` field of `.`.
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Note that the left-hand side of `|=` refers to a value in `.`.
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Thus `$var.foo |= . + 1` won't work as expected (`$var.foo` is
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not a valid or useful path expression in `.`); use `$var |
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@@ -3588,13 +3595,27 @@ sections:
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body: |
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jq has a few operators of the form `a op= b`, which are all
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equivalent to `a |= . op b`. So, `+= 1` can be used to
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increment values, being the same as `|= . + 1`.
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equivalent to `(b as $b | a |= . op $b)`. So, `+= 1` can be
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used to increment values, being the same as `|= . + 1`.
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Note that the right-hand side `b` is evaluated with the same
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input as the left-hand side, not with the value at the path
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being updated. So `.a += .b` adds the value of `.b` to `.a`,
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whereas `.a |= . + .b` would look up `.b` inside the value of
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`.a`.
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If `b` outputs multiple values, then `a op= b` outputs one
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result for each of them.
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examples:
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- program: .foo += 1
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input: '{"foo": 42}'
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output: ['{"foo": 43}']
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- program: '.a += .b'
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input: '{"a": 1, "b": 2}'
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output: ['{"a": 3, "b": 2}']
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- program: '.a += (1,2)'
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input: '{"a": 0}'
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output: ['{"a": 1}', '{"a": 2}']
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- title: "Plain assignment: `=`"
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body: |
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