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PO"is" operator behaving a bit strangely
 

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  1. PO"is" operator behaving a bit strangely
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    <p>1) According to my book, <code>is</code> operator can check whether expression <code>E</code> (<code>E is type</code>) can be converted to the target type only if <code>E</code> is either a reference conversion, boxing or unboxing. Since in the following example <code>is</code> doesn’t check for either of the three types of conversion, the code shouldn’t work, but it does: </p> <pre><code> long l; // EDIT - I forgot to add this line of code in my initial post int i=100; if (i is long) //EDIT - in my initial post I've claimed condition returns true, but it really returns false l = i; </code></pre> <p>2) </p> <p>a) </p> <pre><code> B b; A a = new A(); if (a is B) b = (B)a; int i = b.l; class A { public int l = 100; } class B:A { } </code></pre> <p>The above code always causes compile time error <code>“Use of unassigned variable”</code>. If condition <code>a is B</code> evaluates to <code>false</code>, then <code>b</code> won’t be assigned a value, but if condition is <code>true</code>, then it will. And thus by allowing such a code compiler would have no way of knowing whether the usage of <code>b</code> in code following the <code>if</code> statement is valid or not ( due to not knowing whether <code>a is b</code> evaluates to <code>true</code> or <code>false</code>) , but why should it know that? Intsead why couldn’t runtime handle this? </p> <p>b) But if instead we’re dealing with non reference types, then compiler doesn’t complain, even though the code is identical.Why? </p> <pre><code> int i = 100; long l; if (i is long) l = i; </code></pre> <p>thank you</p>
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    1. COPersonally, I'm surprised that `(i is long) == true`. The behavior of (2) is perfectly straightforward, though.
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      1. USJSBձոգչ
    2. COBTW, C# is case-sensitive. There is no `IS` operator.
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      1. USJohn Saunders
    3. CO@JS Bangs - simple: it isn't.
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      1. USMarc Gravell
 

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