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    <p>It's that way by design in C#, and, in fact, dates back all the way to C/C++ - the latter also promotes operands to <code>int</code>, you just usually don't notice because <code>int -&gt; char</code> conversion there is implicit, while it's not in C#. This doesn't just apply to <code>|</code> either, but to all arithmetic and bitwise operands - e.g. adding two <code>byte</code>s will give you an <code>int</code> as well. I'll quote the relevant part of the spec here:</p> <blockquote> <p>Binary numeric promotion occurs for the operands of the predefined +, –, *, /, %, &amp;, |, ^, ==, !=, >, &lt;, >=, and &lt;= binary operators. Binary numeric promotion implicitly converts both operands to a common type which, in case of the non-relational operators, also becomes the result type of the operation. Binary numeric promotion consists of applying the following rules, in the order they appear here: </p> <ul> <li><p>If either operand is of type decimal, the other operand is converted to type decimal, or a compile-time error occurs if the other operand is of type float or double.</p></li> <li><p>Otherwise, if either operand is of type double, the other operand is converted to type double. </p></li> <li><p>Otherwise, if either operand is of type float, the other operand is converted to type float.</p></li> <li><p>Otherwise, if either operand is of type ulong, the other operand is converted to type ulong, or a compile-time error occurs if the other operand is of type sbyte, short, int, or long.</p></li> <li><p>Otherwise, if either operand is of type long, the other operand is converted to type long.</p></li> <li><p>Otherwise, if either operand is of type uint and the other operand is of type sbyte, short, or int, both operands are converted to type long.</p></li> <li><p>Otherwise, if either operand is of type uint, the other operand is converted to type uint.</p></li> <li><p>Otherwise, both operands are converted to type int.</p></li> </ul> </blockquote> <p>I don't know the exact rationale for this, but I can think about one. For arithmetic operators especially, it might be a bit surprising for people to get <code>(byte)200 + (byte)100</code> suddenly equal to <code>44</code>, even if it makes some sense when one carefully considers the types involved. On the other hand, <code>int</code> is generally considered a type that's "good enough" for arithmetic on most typical numbers, so by promoting both arguments to <code>int</code>, you get a kind of "just works" behavior for most common cases.</p> <p>As to why this logic was also applied to bitwise operators - I imagine this is so mostly for consistency. It results in a single simple rule that is common for <em>all</em> non-boolean binary types.</p> <p>But this is all mostly guessing. Eric Lippert would probably be the one to ask about the real motives behind this decision for C# at least (though it would be a bit boring if the answer is simply "it's how it's done in C/C++ and Java, and it's a good enough rule as it is, so we saw no reason to change it").</p>
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    1. USPavel Minaev
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    1. COThis is really interesting. This post deservses a lot of votes:)
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      1. USStilgar
    2. COsweet answer! good detail! Could've used a code snippet to demonstrate what would fix his problem, but all in all very interesting.
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      1. USBen Lesh
    3. COFor completeness it would be worth mentioning that the `|=` etc operators do a cast implicitly.
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      1. USRoman Starkov
 

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