Re: Documentation bug for __builtin_choose_expr

[Andreas Schwab <schwab at suse dot de>]

Jamie Lokier <jamie@shareable.org> writes:

> 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.

I don't see that.  For me the word "also" implies "another analogousness".
IMHO this is fact is worth noting because the usual rules for function
calls in C is to evaluate all its arguments first, whereas this builtin
does not do that even though it uses a function-like notation.

Andreas.

-- 
Andreas Schwab, SuSE Labs, schwab@suse.de
SuSE Linux Products GmbH, Maxfeldstraße 5, 90409 Nürnberg, Germany
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"And now for something completely different."