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  1. PO
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    <p>Use Dave Anderson's as a basis, but also consider the special cases of where:</p> <pre><code>1) x is 0 2) e is negative. </code></pre> <p>Again, no code so you can try to figure it out yourself :-)</p> <p><strong>Update:</strong> Make sure you create a number of test cases, and you make sure everyone works as you think it should. Once the tests pass, you'll know that your recursive power function is working correctly.</p> <p><strong>Example:</strong> Having had time to figure it out yourself, I though I'd present a solution, with tests:</p> <pre><code>int main(void) { // n = 0 special case test(0, 0, 1); test(4, 0, 1); test(-5, 0, 1); // x = 0 special case test(0, 0, 1); test(0, 2, 0); // normal use test(4, 1, 4); test(4, -1, 0.25); test(-4, 3, -64); test(8, 2, 64); test(2, 3, 8); test(2, -3, 0.125); test(2, -5, 0.03125); // Invalid input tests std::cout &lt;&lt; std::endl &lt;&lt; "Invalid input tests" &lt;&lt; std::endl; test (0, -2, NULL); test(0, -4, NULL); // Negative Tests std::cout &lt;&lt; std::endl &lt;&lt; "Negative tests (expect failure)" &lt;&lt; std::endl; test(4, 0, 4); test(2, 1, 1); test(2, -5, 0.0313); return 0; } double power(int x, int n) { // check for invalid input if (n == 0) { return 1; } if (n &gt; 0) { return x * power(x, n - 1); } else if (n &lt; 0) { return 1 / (x * power(x, -n - 1)); } } bool test(int x, int n, double expected) { if (x == 0 &amp;&amp; n &lt; 0) { std::cout &lt;&lt; "Testing " &lt;&lt; x &lt;&lt; "^" &lt;&lt; n &lt;&lt; ", result = 'Invalid input'." &lt;&lt; std::endl; return false; } double result = power(x, n); std::cout &lt;&lt; "Testing " &lt;&lt; x &lt;&lt; "^" &lt;&lt; n &lt;&lt; ", result = " &lt;&lt; result &lt;&lt; ". Expected " &lt;&lt; expected &lt;&lt; " - test " &lt;&lt; ((result == expected) ? "PASSED" : "FAILED") &lt;&lt; std::endl; return true; } </code></pre> <p><strong>Output:</strong></p> <pre><code>Testing 0^0, result = 1. Expected 1 - test PASSED Testing 4^0, result = 1. Expected 1 - test PASSED Testing -5^0, result = 1. Expected 1 - test PASSED Testing 0^0, result = 1. Expected 1 - test PASSED Testing 0^2, result = 0. Expected 0 - test PASSED Testing 4^1, result = 4. Expected 4 - test PASSED Testing 4^-1, result = 0.25. Expected 0.25 - test PASSED Testing -4^3, result = -64. Expected -64 - test PASSED Testing 8^2, result = 64. Expected 64 - test PASSED Testing 2^3, result = 8. Expected 8 - test PASSED Testing 2^-3, result = 0.125. Expected 0.125 - test PASSED Testing 2^-5, result = 0.03125. Expected 0.03125 - test PASSED Invalid input tests Testing 0^-2, result = 'Invalid input'. Testing 0^-4, result = 'Invalid input'. Negative tests (expect failure) Testing 4^0, result = 1. Expected 4 - test FAILED Testing 2^1, result = 2. Expected 1 - test FAILED Testing 2^-5, result = 0.03125. Expected 0.0313 - test FAILED </code></pre>
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    1. POMultiplying to powers using recursion
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    1. COthe first case you give should be accounted for, simply by the fact that anything you give it will just be multiplied by 0. however, the negative one is a bit trickier, and was the biggest problem I thought would come up. perhaps using an if(x>0) {dave anderson code}; if(x<0) {similar, but divide instead of multiply, and add towards 0}. Think I'm on the right track?
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    2. CO1) Could be made more efficient as a special case, and avoid any recursion. 2) For each recursion, `e` needs to *move towards* `1`. 3) Dave's example will always return `1` for positive powers, how might you stop that?
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    3. COYou are almost correct but it's the `if (e > 0) {...}; if (e < 0) {...}` i.e. `e` rather than `x`.
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