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    <p>I like the answer of R. slightly tweaked:</p> <pre><code>float f = 1234.56789; printf("%d.%.0f", f, 1000*(f-(int)f)); </code></pre> <p>'1000' determines the precision.</p> <p><em>Power to the 0.5 rounding.</em></p> <p><strong>EDIT</strong></p> <p>Ok, this answer was edited a few times and I lost track what I was thinking a few years back (and originally it did not fill all the criteria). So here is a new version (that fills all criteria and handles negative numbers correctly):</p> <pre><code>double f = 1234.05678900; char s[100]; int decimals = 10; sprintf(s,"%.*g", decimals, ((int)(pow(10, decimals)*(fabs(f) - abs((int)f)) +0.5))/pow(10,decimals)); printf("10 decimals: %d%s\n", (int)f, s+1); </code></pre> <p>And the test cases:</p> <pre><code>#import &lt;stdio.h&gt; #import &lt;stdlib.h&gt; #import &lt;math.h&gt; int main(void){ double f = 1234.05678900; char s[100]; int decimals; decimals = 10; sprintf(s,"%.*g", decimals, ((int)(pow(10, decimals)*(fabs(f) - abs((int)f)) +0.5))/pow(10,decimals)); printf("10 decimals: %d%s\n", (int)f, s+1); decimals = 3; sprintf(s,"%.*g", decimals, ((int)(pow(10, decimals)*(fabs(f) - abs((int)f)) +0.5))/pow(10,decimals)); printf(" 3 decimals: %d%s\n", (int)f, s+1); f = -f; decimals = 10; sprintf(s,"%.*g", decimals, ((int)(pow(10, decimals)*(fabs(f) - abs((int)f)) +0.5))/pow(10,decimals)); printf(" negative 10: %d%s\n", (int)f, s+1); decimals = 3; sprintf(s,"%.*g", decimals, ((int)(pow(10, decimals)*(fabs(f) - abs((int)f)) +0.5))/pow(10,decimals)); printf(" negative 3: %d%s\n", (int)f, s+1); decimals = 2; f = 1.012; sprintf(s,"%.*g", decimals, ((int)(pow(10, decimals)*(fabs(f) - abs((int)f)) +0.5))/pow(10,decimals)); printf(" additional : %d%s\n", (int)f, s+1); return 0; } </code></pre> <p>And the output of the tests:</p> <pre><code> 10 decimals: 1234.056789 3 decimals: 1234.057 negative 10: -1234.056789 negative 3: -1234.057 additional : 1.01 </code></pre> <p>Now, all criteria are met:</p> <ul> <li>maximum number of decimals behind the zero is fixed</li> <li>trailing zeros are removed</li> <li>it does it mathematically right (right?)</li> <li>works (now) also when first decimal is zero </li> </ul> <p>Unfortunately this answer is a two-liner as <code>sprintf</code> does not return the string.</p>
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    1. POAvoid trailing zeroes in printf()
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    1. CO+1 For making this a one-liner.
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    2. COThis doesn't actually seem to trim trailing zeroes, though? It can gladly spit out something like 1.1000.
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    3. COThe question and my answer are not the originals anymore... I will check later.
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