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POWhy use only the lower five bits of the shift operand when shifting a 32-bit value? (e.g. (UInt32)1 << 33 == 2)
 

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  1. POWhy use only the lower five bits of the shift operand when shifting a 32-bit value? (e.g. (UInt32)1 << 33 == 2)
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    <p>Consider the following code:</p> <pre><code>UInt32 val = 1; UInt32 shift31 = val &lt;&lt; 31; // shift31 == 0x80000000 UInt32 shift32 = val &lt;&lt; 32; // shift32 == 0x00000001 UInt32 shift33 = val &lt;&lt; 33; // shift33 == 0x00000002 UInt32 shift33a = (UInt32)((UInt64)val &lt;&lt; 33); // shift33a == 0x00000000 </code></pre> <p>It doesn't generate a warning (about using a shift greater than 32) so it must be an expected behavior.</p> <p>The code that actually gets put out to the generated assembly (or at least Reflector's interpretation of the code) is</p> <pre><code> uint val = 1; uint shift31 = val &lt;&lt; 0x1f; uint shift32 = val; uint shift33 = val &lt;&lt; 1; uint shift33a = val &lt;&lt; 0x21; </code></pre> <p>The IL (again, using Reflector) is</p> <pre><code>L_0000: nop L_0001: ldc.i4.1 L_0002: stloc.0 L_0003: ldloc.0 L_0004: ldc.i4.s 0x1f L_0006: shl L_0007: stloc.1 L_0008: ldloc.0 L_0009: stloc.2 L_000a: ldloc.0 L_000b: ldc.i4.1 L_000c: shl L_000d: stloc.3 L_000e: ldloc.0 L_000f: conv.u8 L_0010: ldc.i4.s 0x21 L_0012: shl L_0013: conv.u4 L_0014: stloc.s shift33a </code></pre> <p>I understand <strong>what</strong> is going on (it's described in <a href="http://msdn.microsoft.com/en-us/library/aa691377.aspx" rel="noreferrer">MSDN</a>); when the code is compiled, only the lower 5 bits are being used when shifting a 32-bit value... I'm curious as to <strong>why</strong> this happens. </p> <p>(The way <code>shift33a</code> comes out also makes me think that something isn't quite right with Reflector, as their c# presentation of the IL will compile to something different)</p> <p><strong>The question(s):</strong></p> <ul> <li><strong>Why</strong> are only the lower 5 bits of "the value to shift by" used? </li> <li>If "it doesn't make sense to shift more than 31 bits", why isn't there a warning?</li> <li>Is this a backwards compatilbility thing (i.e. is this what programmers "expect" to happen)?</li> <li>Am I correct that the underlying IL can do shifts of more than 31 bits (as in <code>L_0010: ldc.i4.s 0x21</code>) but the compiler is trimming the values?</li> </ul>
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    1. USDaniel LeCheminant
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    1. CONot even the specification gives insight on the "why". You may try asking C# compiler guys like Eric Lippert.
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      1. USJoey
    2. CO@Johannes: Yeah, I was just wondering if someone had a use case or something for why it makes sense to behave this way :-/
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      1. USDaniel LeCheminant
    3. COYou should never trust Reflector. I have spent hours chasing bugs, because I trusted the output. The IL is normally the most reliable, but I have had occasions that even didn't go well. Try ILDASM instead.
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