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PODoes an observable difference exist using `unsigned long` and `unsigned int` in C (or C++) when both are 32 bits wide?
 

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  1. PODoes an observable difference exist using `unsigned long` and `unsigned int` in C (or C++) when both are 32 bits wide?
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    <p>I'm using an MPC56XX (embedded systems) with a compiler for which an <code>int</code> and a <code>long</code> are both 32 bits wide.</p> <p>In a required software package we had the following definitions for 32-bit wide types:</p> <pre><code>typedef signed int sint32; typedef unsigned int uint32; </code></pre> <p>In a new release this was changed without much documentation to:</p> <pre><code>typedef signed long sint32; typedef unsigned long uint32; </code></pre> <p>I can see why this would be a good thing: Integers have a conversion rank between <code>short</code> and <code>long</code>, so theoretically extra conversions can apply when using the first set of definitions.</p> <p>My question: Given the above change forced upon us by the package authors, is there a situation imaginable where such a change would change the compiled code, correctly leading to a different result?</p> <p>I'm familiar with the "usual unary conversions" and the "usual binary conversions", but I have a hard time coming up with a concrete situation where this could really ruin my existing code. But is it really irrelevant?</p> <p>I'm currently working in a pure C environment, using C89/C94, but I'd be interested in both C and C++ issues.</p> <p>EDIT: I know that mixing <code>int</code> with <code>sint32</code> may produce different results when it's redefined. But we're not allowed to use the original C types directly, only the typedef'ed ones.<br> I'm looking for a sample (expression or snippet) using constants, unary/binary operators, casts, etc. with a different but correct compilation result based on the changed type definition.</p>
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    1. USJohan Bezem
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    1. COThey are still different types
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      1. PODoes an observable difference exist using `unsigned long` and `unsigned int` in C (or C++) when both are 32 bits wide?
      1. USJohannes Schaub - litb
    2. CO¤ As @JohannesSchaub-litb notes in his comment, they are different types. That means you can different results from overload resolution, you get different `typeid` results, etc. However, regarding arithmetic operations it's the same, unless you have really really really perverse in-house compiler. Cheers & hth.,
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      1. PODoes an observable difference exist using `unsigned long` and `unsigned int` in C (or C++) when both are 32 bits wide?
      1. USCheers and hth. - Alf
    3. CO@AlfP.Steinbach Yes, that's what I initially thought. But I have a feeling that a similar situation like in [another of my questions](http://stackoverflow.com/q/8069516/1029106) might exist. I currently cannot think of any, but in SO there's many smarter people than me, I'd rather ask.
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      1. USJohan Bezem
 

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