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Documentation bug for __builtin_choose_expr
- From: Jamie Lokier <jamie at shareable dot org>
- To: gcc at gcc dot gnu dot org
- Date: Mon, 22 Nov 2004 04:26:52 +0000
- Subject: Documentation bug for __builtin_choose_expr
The documentation for __builtin_choose_expr says:
-- Built-in Function: TYPE __builtin_choose_expr (CONST_EXP, EXP1, EXP2)
You can use the built-in function `__builtin_choose_expr' to
evaluate code depending on the value of a constant expression.
This built-in function returns EXP1 if CONST_EXP, which is a
constant expression that must be able to be determined at compile
time, is nonzero. Otherwise it returns 0.
This built-in function is analogous to the `? :' operator in C,
except that the expression returned has its type unaltered by
promotion rules. Also, the built-in function does not evaluate
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
the expression that was not chosen. For example, if CONST_EXP
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
evaluates to true, EXP2 is not evaluated even if it has
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
side-effects.
^^^^^^^^^^^^^
The way this is written implies that the underscored behaviour is
different from the `? :' operator in C.
But the `? :' operator does not evaluate the expression that was not
chosen either. If the condition evaluates to true, EXP2 is not
evaluated in `COND ? EXP1 : EXP2'.
It would be better to replace that paragraph with:
This built-in function is analogous to the `? :' operator in C,
except that the expression returned has its type unaltered by
promotion rules.
If there is some other effect of __builtin_choose_expr, for example if
it were to inhibit compilation or warnings in the unselected branch in
some way, it would be useful to mention that. I don't know of any
such effects. (Inhibiting warnings in the unselected branch and
promising to not expand large inline functions in that branch would be
a useful addition imho).
-- Jamie