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copied!<p>This can't be done with the normal <code>printf</code> format specifiers. The closest you could get would be:</p> <pre><code>printf("%.6g", 359.013); // 359.013 printf("%.6g", 359.01); // 359.01 </code></pre> <p>but the ".6" is the <em>total</em> numeric width so</p> <pre><code>printf("%.6g", 3.01357); // 3.01357 </code></pre> <p>breaks it.</p> <p>What you <em>can</em> do is to <code>sprintf("%.20g")</code> the number to a string buffer then manipulate the string to only have N characters past the decimal point.</p> <p>Assuming your number is in the variable num, the following function will remove all but the first <code>N</code> decimals, then strip off the trailing zeros (and decimal point if they were all zeros).</p> <pre><code>char str[50]; sprintf (str,"%.20g",num); // Make the number. morphNumericString (str, 3); : : void morphNumericString (char *s, int n) { char *p; int count; p = strchr (s,'.'); // Find decimal point, if any. if (p != NULL) { count = n; // Adjust for more or less decimals. while (count >= 0) { // Maximum decimals allowed. count--; if (*p == '\0') // If there's less than desired. break; p++; // Next character. } *p-- = '\0'; // Truncate string. while (*p == '0') // Remove trailing zeros. *p-- = '\0'; if (*p == '.') { // If all decimals were zeros, remove ".". *p = '\0'; } } } </code></pre> <hr> <p>If you're not happy with the truncation aspect (which would turn <code>0.12399</code> into <code>0.123</code> rather than rounding it to <code>0.124</code>), you can actually use the rounding facilities already provided by <code>printf</code>. You just need to analyse the number before-hand to dynamically create the widths, then use those to turn the number into a string:</p> <pre><code>#include <stdio.h> void nDecimals (char *s, double d, int n) { int sz; double d2; // Allow for negative. d2 = (d >= 0) ? d : -d; sz = (d >= 0) ? 0 : 1; // Add one for each whole digit (0.xx special case). if (d2 < 1) sz++; while (d2 >= 1) { d2 /= 10.0; sz++; } // Adjust for decimal point and fractionals. sz += 1 + n; // Create format string then use it. sprintf (s, "%*.*f", sz, n, d); } int main (void) { char str[50]; double num[] = { 40, 359.01335, -359.00999, 359.01, 3.01357, 0.111111111, 1.1223344 }; for (int i = 0; i < sizeof(num)/sizeof(*num); i++) { nDecimals (str, num[i], 3); printf ("%30.20f -> %s\n", num[i], str); } return 0; } </code></pre> <p>The whole point of <code>nDecimals()</code> in this case is to correctly work out the field widths, then format the number using a format string based on that. The test harness <code>main()</code> shows this in action:</p> <pre><code> 40.00000000000000000000 -> 40.000 359.01335000000000263753 -> 359.013 -359.00999000000001615263 -> -359.010 359.00999999999999090505 -> 359.010 3.01357000000000008200 -> 3.014 0.11111111099999999852 -> 0.111 1.12233439999999995429 -> 1.122 </code></pre> <p>Once you have the correctly rounded value, you can once again pass that to <code>morphNumericString()</code> to remove trailing zeros by simply changing:</p> <pre><code>nDecimals (str, num[i], 3); </code></pre> <p>into:</p> <pre><code>nDecimals (str, num[i], 3); morphNumericString (str, 3); </code></pre> <p>(or calling <code>morphNumericString</code> at the end of <code>nDecimals</code> but, in that case, I'd probably just combine the two into one function), and you end up with:</p> <pre><code> 40.00000000000000000000 -> 40 359.01335000000000263753 -> 359.013 -359.00999000000001615263 -> -359.01 359.00999999999999090505 -> 359.01 3.01357000000000008200 -> 3.014 0.11111111099999999852 -> 0.111 1.12233439999999995429 -> 1.122 </code></pre>

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