sergiotarxz/nuttx
1
0
Fork 0
Code Issues Pull Requests Releases Wiki Activity

Documentation/NuttXCCodingStandard.html: Another small change discouraging the practice of enclosing the value argument of 'return' statements in parentheses.

Browse Source
This commit is contained in:
Gregory Nutt 2019-07-06 13:08:27 -06:00
parent bd248b52e4
commit 9be93c710a
1 changed files with 28 additions and 21 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

49
Documentation/NuttXCCodingStandard.html
View File

@ -87,7 +87,7 @@
<h1><big><font color="#3c34ec">
<i>NuttX C Coding Standard</i>
</font></big></h1>
<p>Last Updated: July 5, 2019</p>
<p>Last Updated: July 6, 2019</p>
</td>
</tr>
</table>
@ -2092,7 +2092,14 @@ ptr = (FAR struct somestruct_s *)value;
As a rule of thumb, the length of a function should be limited so that it would fit on a single page (if you were to print the source code).
</li>
<li>
<b>Space after the function body</b>
<b>Return Statement</b>.
The argument of the <code>return</code> statement should <i>not</i> be enclosed in parentheses.
A reasonable exception is the case where the returned value argument is a complex expression and where the parentheses improve the readability of the code.
Such complex expressions might be Boolean expressions or expressions containing conditions.
Simple arithmetic computations would not be considered <i>complex</i> expressions.
</li>
<li>
<b>Space after the function body</b>.
A one (and only one) blank line must follow the closing right brace of the function body.
</li>
</ul>
@ -2121,7 +2128,7 @@ ptr = (FAR struct somestruct_s *)value;
}
}
return e / a;
return (e / a);
}
</ul></pre></font>
</td></tr>
@ -2129,28 +2136,28 @@ ptr = (FAR struct somestruct_s *)value;
<p><font color="green"><b>Correct</b></p>
<ul><pre>
int myfunction(int a, int b)
{
int c;
int d;
int e;
int i;
{
int c;
int d;
int e;
int i;
c = a
d = b;
e = c + d;
c = a
d = b;
e = c + d;
for (i = 0; i &lt; a; i++)
{
int j;
for (i = 0; i &lt; a; i++)
{
int j;
for (j = 0; j &lt; b; j++)
{
e += j * d;
}
}
for (j = 0; j &lt; b; j++)
{
e += j * d;
}
}
return e / a;
}
return e / a;
}
</ul></pre></font>
</td></tr>
</table></center>