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<html xmlns="http://www.w3.org/1999/xhtml"><head><meta http-equiv="Content-Type" content="text/html; charset=utf-8" /><title>Hyperlinked ECMA C# Language Specification</title><meta name="author" content="Jon Jagger" /><link rel="stylesheet" href="ecma334.css"></link></head><body><div align="right"><em><a href="http://www.jaggersoft.com">Jon Jagger</a></em></div><div align="right"><a href="mailto:jon@jaggersoft.com">jon@jaggersoft.com</a></div><form method="get" action="http://search.atomz.com/search/"><input size="30" name="sp-q"></input><input type="submit" value="Search C# Spec"></input><input type="hidden" name="sp-a" value="sp10024177"></input><input type="hidden" name="sp-f" value="ISO-8859-1"></input></form><a href="toc.htm">Table of Contents</a> <a href="1.htm">1</a> <a href="2.htm">2</a> <a href="3.htm">3</a> <a href="4.htm">4</a> <a href="5.htm">5</a> <a href="6.htm">6</a> <a href="7.htm">7</a> <a href="8.htm">8</a> <a href="9.htm">9</a> <a href="10.htm">10</a> <a href="11.htm">11</a> <a href="12.htm">12</a> <a href="13.htm">13</a> <a href="14.htm">14</a> <a href="15.htm">15</a> <a href="16.htm">16</a> <a href="17.htm">17</a> <a href="18.htm">18</a> <a href="19.htm">19</a> <a href="20.htm">20</a> <a href="21.htm">21</a> <a href="22.htm">22</a> <a href="23.htm">23</a> <a href="24.htm">24</a> <a href="25.htm">25</a> <a href="notes.htm">Notes</a> <a href="HyperlinkedCSharpECMA.zip">Download</a><span class="ruler"></span><span class="heading">ECMA-334 C# Language Specification</span><span class="navigate"><a href="14.13.1.htm"><img src="previous.gif" alt="previous" border="0" /></a><a href="14.13.3.htm"><img src="next.gif" alt="next" border="0" /></a></span><span class="clause-depth"><a href="13.htm"><img src="previous.gif" alt="previous at this level" border="0" /></a><a href="15.htm"><img src="next.gif" alt="next at this level" border="0" /></a> <span class="clause-number-link"><a href="14.htm">14</a></span><span class="clause-title-previous"> Expressions</span></span><span class="clause-depth"><a href="14.12.htm"><img src="previous.gif" alt="previous at this level" border="0" /></a><a href="14.14.htm"><img src="next.gif" alt="next at this level" border="0" /></a> <span class="clause-number-link"><a href="14.13.htm">14.13</a></span><span class="clause-title-previous"> Assignment operators</span></span><span class="clause-depth"><a href="14.13.1.htm"><img src="previous.gif" alt="previous at this level" border="0" /></a><a href="14.13.3.htm"><img src="next.gif" alt="next at this level" border="0" /></a> <span class="clause-number">14.13.2</span><span class="clause-title"> Compound assignment</span></span><span class="locator">
Paragraph 1</span><span class="paragraph"><span class="sentence"><span class="sentence-number">1</span> <a name="P1S1"></a>An operation of the form x op= y is processed by applying binary operator overload resolution (<a href="14.2.4.htm">§14.2.4</a>) as if the operation was written x op y.</span> <span class="sentence"><span class="sentence-number">2</span> <a name="P1S2"></a>Then, </span><ul><li><span class="sentence"><span class="sentence-number">3</span> <a name="P1S3"></a> If the return type of the selected operator is implicitly convertible to the type of x, the operation is evaluated as x = x op y, except that x is evaluated only once.</span> </li><li><span class="sentence"><span class="sentence-number">4</span> <a name="P1S4"></a> Otherwise, if the selected operator is a predefined operator, if the return type of the selected operator is explicitly convertible to the type of x, and if y is implicitly convertible to the type of x, then the operation is evaluated as x = (T)(x op y), where T is the type of x, except that x is evaluated only once.</span> </li><li><span class="sentence"><span class="sentence-number">5</span> <a name="P1S5"></a> Otherwise, the compound assignment is invalid, and a compile-time error occurs.</span> </li></ul></span><span class="locator">
Paragraph 2</span><span class="paragraph"><span class="sentence"><span class="sentence-number">1</span> <a name="P2S1"></a>The term "evaluated only once" means that in the evaluation of x op y, the results of any constituent expressions of x are temporarily saved and then reused when performing the assignment to x.</span> <span class="example">[Example: For example, in the assignment A()[B()] += C(), where A is a method returning int[], and B and C are methods returning <span class="keyword">int</span>, the methods are invoked only once, in the order A, B, C. end example]</span> </span><span class="locator">
Paragraph 3</span><span class="paragraph"><span class="sentence"><span class="sentence-number">1</span> <a name="P3S1"></a>When the left operand of a compound assignment is a property access or indexer access, the property or indexer must have both a get accessor and a set accessor.</span> <span class="sentence"><span class="sentence-number">2</span> <a name="P3S2"></a>If this is not the case, a compile-time error occurs.</span> </span><span class="locator">
Paragraph 4</span><span class="paragraph"><span class="sentence"><span class="sentence-number">1</span> <a name="P4S1"></a>The second rule above permits x op= y to be evaluated as x = (T)(x op y) in certain contexts.</span> <span class="sentence"><span class="sentence-number">2</span> <a name="P4S2"></a>The rule exists such that the predefined operators can be used as compound operators when the left operand is of type <span class="keyword">sbyte</span>, <span class="keyword">byte</span>, <span class="keyword">short</span>, <span class="keyword">ushort</span>, or <span class="keyword">char</span>.</span> <span class="sentence"><span class="sentence-number">3</span> <a name="P4S3"></a>Even when both arguments are of one of those types, the predefined operators produce a result of type <span class="keyword">int</span>, as described in <a href="14.2.6.2.htm">§14.2.6.2</a>.</span> <span class="sentence"><span class="sentence-number">4</span> <a name="P4S4"></a>Thus, without a cast it would not be possible to assign the result to the left operand.</span> </span><span class="locator">
Paragraph 5</span><span class="paragraph"><span class="sentence"><span class="sentence-number">1</span> <a name="P5S1"></a>The intuitive effect of the rule for predefined operators is simply that x op= y is permitted if both of x op y and x = y are permitted.</span> <span class="example">[Example: In the example <pre class="code-example">
byte b = 0;
char ch = '\0';
int i = 0;
b += 1; // Ok
b += 1000; // Error, b = 1000 not permitted
b += i; // Error, b = i not permitted
b += (byte)i; // Ok
ch += 1; // Error, ch = 1 not permitted
ch += (char)1; // Ok
</pre>the intuitive reason for each error is that a corresponding simple assignment would also have been an error. end example]</span> </span><span class="ruler"></span><table><tr><td><table align="left" bgcolor="navy"><tr bgcolor="navy"><td><font face="Arial,sans-serif" size="6" color="yellow"><strong>{ JSL }</strong></font></td></tr></table></td></tr><tr><td><font face="Arial,sans-serif" size="2" color="navy"><strong>Jagger Software Ltd</strong></font></td></tr><tr><td><font face="Arial,sans-serif" size="2" color="navy"><strong>Company # 4070126</strong></font></td></tr><tr><td><font face="Arial,sans-serif" size="2" color="navy"><strong>VAT # 762 5213 42</strong></font></td></tr></table><img src="valid-html401.png" align="left" height="31" width="88" alt="Valid HTML 4.01" /><img src="vcss.gif" align="left" height="31" width="88" alt="Valid CSS" /></body></html>⌨️ 快捷键说明
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