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Don't allow function values in assert tests.

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Currently you can write
  assert(() { ... });
and the function will be called and the return value used as the assert test.
This feature isn't really worth its own complexity - if you want to get the same effect, you can just write all the function:
  assert(() { ... }());
With asserts in const initializer lists, where the function call is not possible anyway, the feature went from being not very useful to being actual an complication and exception for users to remember.

R=eernst@google.com, rnystrom@google.com

Review-Url: https://codereview.chromium.org/2974763002 .
This commit is contained in:
Lasse Reichstein Holst Nielsen 2017-08-10 15:29:16 +02:00
parent 3803057169
commit e329e1ce29
1 changed files with 13 additions and 17 deletions
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30
docs/language/dartLangSpec.tex
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@ -7,9 +7,9 @@
\usepackage{lmodern}
\usepackage[T1]{fontenc}
\newcommand{\code}[1]{{\sf #1}}
\title{Dart Programming Language Specification \\
\title{Dart Programming Language Specification \\
{5th edition draft}\\
{\large Version 1.15.0}}
{\large Version 2.0.0-dev}}
% For information about Location Markers (and in particular the
% commands \LMHash and \LMLabel), see the long comment at the
@ -19,6 +19,9 @@
% =======
% Significant changes to the specification.
%
% 2.0
% - Don't allow functions as assert test values.
%
% 1.15
% - Change how language specification describes control flow.
% - Object initialization now specifies initialization order correctly.
@ -6757,41 +6760,34 @@ An {\em assert statement} is used to disrupt normal execution if a given boolean
\LMHash{}
The grammar allows a trailing comma before the closing parenthesis,
similarly to an argument list. That comma, if present, has no effect.
An assertion with a trailing comme is equivalent to one with that
An assertion with a trailing comma is equivalent to one with that
comma removed.
\LMHash{}
An assertion on the form \code{\ASSERT($e$))} is equivalent to an assertion
on the form \code{\ASSERT($e$, null)}.
An assertion on the form \code{\ASSERT($e$))} is equivalent to an assertion on the form \code{\ASSERT($e$, null)}.
\LMHash{}
Execution of an assert statement executes the assertion as described below
and completes in the same way as the assertion.
\LMHash{}
In production mode the assertion has no effect
In production mode an assertion has no effect
and its execution immediately completes normally (\ref{completion}).
In checked mode,
execution of an assertion \code{\ASSERT{}($c$, $e$)} proceeds as follows:
\LMHash{}
The expression $c$ is evaluated to an object $o$.
If the class of $o$ is a subtype of \code{Function}
and the assertion does not occur in the initializer list of a \CONST{} constructor,
then let $r$ be the result of invoking $o$ with no arguments.
Otherwise let $r$ be $o$.
The expression $c$ is evaluated to an object $r$.
It is a dynamic type error if $r$ is not of type \code{bool}.
\commentary{Hence it is a compile-time error if that situation arises during evaluation of an assertion in a \CONST{} constructor invocation.}
If $r$ is \TRUE{}, execution of the assert statement completes normally (\ref{completion}).
If $r$ is \TRUE{} then execution of the assert statement completes normally (\ref{completion}).
Otherwise, $e$ is evaluated to an object $m$
and then the execution of the assert statement throws (\ref{completion}) an \code{AssertionError} containing $m$ and with a stack trace corresponding to the current execution state at the \ASSERT{} statement.
and then the execution of the assert statement throws (\ref{completion}) an \code{AssertionError} containing $m$ and with a stack trace corresponding to the current execution state at the assertion.
\LMHash{}
It is a static type warning if the type of $e$ may not be assigned to either
\code{bool} or $() \rightarrow$ \code{bool}.
If the assertion occurs in a \CONST{} constructor initializer list, it is a static type warning if the type of $e$ may not be assigned to \code{bool}.
It is a static type warning if the type of $e$ may not be assigned to \code{bool}.
\rationale{Why is this a statement, not a built in function call? Because it is handled magically so it has no effect and no overhead in production mode. Also, in the absence of final methods. one could not prevent it being overridden (though there is no real harm in that). It cannot be viewed as a function call that is being optimized away because the argument might have side effects.
\rationale{Why is this a statement, not a built in function call? Because it is handled magically so it has no effect and no overhead in production mode. Also, in the absence of final methods, one could not prevent it being overridden (though there is no real harm in that). It cannot be viewed as a function call that is being optimized away because the arguments might have side effects.
}
%If a lexically visible declaration named \code{assert} is in scope, an assert statement