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    <p>Whether <em>any</em> code construct produces a warning is up to the cleverness of the compiler and the choices made by its authors.</p> <pre><code>char c=9999; </code></pre> <p><code>9999</code> is a constant expression. The compiler can determine, just by analyzing the declaration with no additional context, that it's going to overflow. (Presumably plain <code>char</code> is signed; if it's unsigned, the conversion is well defined -- but a compiler could still choose to warn about it.)</p> <pre><code>int i=9999; char c=i; </code></pre> <p>This has the same semantics, but for a compiler to warn about the initialization of <code>c</code>, it would have to know that <code>i</code> has the value <code>9999</code> (or at least a value outside the range of <code>char</code>) when it analyzes that declaration. Suppose you instead wrote:</p> <pre><code>int i = 9999; i = 42; char c = i; </code></pre> <p>Then clearly no warning would be necessary or appropriate.</p> <p>As <a href="https://stackoverflow.com/a/10503469/827263">James McNellis's answer</a> indicates, a suffiently clever compiler <em>can</em> warn about either case if it performs additional analysis of what's going to happen during the execution of the program. For some compiler, it helps to enable optimization, since the analysis required to optimize code (without breaking it) can also reveal this kind of potential run-time error.</p> <p>I'll note that this is an answer to the question you asked: why is there no warning. The answer you accepted is to the <em>implied</em> question: "I <em>want</em> a warning here; how can I enable it?". I'm not complaining, just observing.</p>
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    1. POwhy no overflow warning when converting int to char
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    1. USKeith Thompson
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    1. CO+1. **pmg** too admitted the comment about one being an initialization and other being an assignment was wrong.Now the whole thing is very clear.Thanks for the time it took to come up with 2 fine answers.
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      1. USRüppell's Vulture
 

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