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precedence reversed

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Mark Pilgrim
2009-09-23 13:31:48 -04:00
parent b7ef6d1c7a
commit c19c32c2eb
1 changed files with 2 additions and 2 deletions
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porting-code-to-python-3-with-2to3.html
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@@ -175,8 +175,8 @@ h3:before{counter-increment:h3;content:'A.' counter(h2) '.' counter(h3) '. '}
<ol>
<li>The simplest form.
<li>The <code>or</code> operator takes precedence over the <code>in</code> operator, so there is no need for parentheses around <code>(x in a_dictionary)</code> or <code>(y in a_dictionary)</code>.
<li>On the other hand, you <em>do</em> need parentheses around <code>(x or y)</code> here, for the same reason&nbsp;&mdash;&nbsp;<code>or</code> takes precedence over <code>in</code>. (Note: this code is completely different from the previous line. Python interprets <code>x or y</code> first, which results in either <var>x</var> (if <var>x</var> is <a href=native-datatypes.html#booleans>true in a boolean context</a>) or <var>y</var>. Then it takes that singular value and checks whether it is a key in <var>a_dictionary</var>.)
<li>The <code>in</code> operator takes precedence over the <code>or</code> operator, so there is no need for parentheses around <code>(x in a_dictionary)</code> or around <code>(y in a_dictionary)</code>.
<li>On the other hand, you <em>do</em> need parentheses around <code>(x or y)</code> here, for the same reason&nbsp;&mdash;&nbsp;<code>in</code> takes precedence over <code>or</code>. (Note: this code is completely different from the previous line. Python interprets <code>x or y</code> first, which results in either <var>x</var> (if <var>x</var> is <a href=native-datatypes.html#booleans>true in a boolean context</a>) or <var>y</var>. Then it takes that singular value and checks whether it is a key in <var>a_dictionary</var>.)
<li>The <code>in</code> operator takes precedence over the <code>+</code> operator, so this form technically doesn&#8217;t need parentheses around <code>(x + y)</code>, but <code>2to3</code> includes them anyway.
<li>This form definitely needs parentheses around <code>(y in a_dictionary)</code>, since the <code>in</code> operator takes precedence over the <code>+</code> operator.
</ol>