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    <p>For big X values (around 700 and above), you'll hit the range limit for doubles (10^308) and cause an infinite loop. You can't do much about it, you should either limit X input range or use some big number library to have extended range. </p> <p>Another workaround is to add this to your loop:</p> <pre><code>if (sum &gt; 1E305) { // we'll most likely run into an infinite loop break; } </code></pre> <p>Note you should handle this case outside the loop afterwards to avoid printing a very large incorrect result.</p> <p>I can't reproduce the problem for <code>0.00000000001</code>, this just returns 1 for me. Negative values run fine, too, although the result is wrong which seems to be an error/limitation in the algorithm. EDIT: To correct this, we can use the fact that <code>e^-x</code> is the same as <code>1 / e^x</code>.</p> <p>Code:</p> <pre><code>#include &lt;stdio.h&gt; double CalcExp(double x){ double eps = 0.0000000000000000001; double elem = 1.0; double sum = 0.0; bool negative = false; int i = 1; sum = 0.0; if (x &lt; 0) { negative = true; x = -x; } do { sum += elem; elem *= x / i; i++; if (sum &gt; 1E305) break; } while (elem &gt;= eps); if (sum &gt; 1E305) { // TODO: Handle large input case here } if (negative) { return 1.0 / sum; } else { return sum; } } int main() { printf("%e\n", CalcExp(0.00000000001)); // Output: 1.000000e+000 printf("%e\n", CalcExp(-4)); // Output: 1.831564e-002 printf("%e\n", CalcExp(-45)); // Output: 2.862519e-020 printf("%e\n", CalcExp(1)); // Output: 2.718282e+000 printf("%e\n", CalcExp(750)); // Output: 1.375604e+305 printf("%e\n", CalcExp(7500000)); // Output: 1.058503e+305 printf("%e\n", CalcExp(-450000)); // Output: 9.241336e-308 return 0; } </code></pre>
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    1. COWow! Thank you very much again! This code is working perfect. You saved me :)
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