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POhow to calculate (a times b) divided by c only using 32-bit integer types even if a times b would not fit such a type
 

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  1. POhow to calculate (a times b) divided by c only using 32-bit integer types even if a times b would not fit such a type
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    <p>Consider the following as a reference implementation:</p> <pre><code>/* calculates (a * b) / c */ uint32_t muldiv(uint32_t a, uint32_t b, uint32_t c) { uint64_t x = a; x = x * b; x = x / c; return x; } </code></pre> <p>I am interested in an implementation (in C or pseudocode) that does not require a 64-bit integer type.</p> <p>I started sketching an implementation that outlines like this:</p> <pre><code>/* calculates (a * b) / c */ uint32_t muldiv(uint32_t a, uint32_t b, uint32_t c) { uint32_t d1, d2, d1d2; d1 = (1 &lt;&lt; 10); d2 = (1 &lt;&lt; 10); d1d2 = (1 &lt;&lt; 20); /* d1 * d2 */ return ((a / d1) * (b /d2)) / (c / d1d2); } </code></pre> <p>But the difficulty is to pick values for d1 and d2 that manage to avoid the overflow ((a / d1) * (b / d2) &lt;= UINT32_MAX) and minimize the error of the whole calculation.</p> <p>Any thoughts?</p>
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    1. USPedro Pedruzzi
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      1. VTUpMod
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    1. COImplementing a 64 bit multiplication and division from the 32 bit one is one obvious way. It
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      1. USAProgrammer
    2. COWhat do you want to happen when the result does not fit in 32 bits? (`UINT_MAX * UINT_MAX / 1` for example)
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      1. USpmg
    3. CO@pmg: same as the reference implementation, surely - return the mathematical result modulo (UINT_MAX+1). That's what "reference implementation" means in my dictionary ;-)
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      1. USSteve Jessop
 

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