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Why Does Integer Math with `pow()` Produce Incorrect Results in C ?

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2024-12-04 22:58:11552browse

Why Does Integer Math with `pow()` Produce Incorrect Results in C  ?

Integer Math with pow() Leading to Incorrect Results

Consider the following code snippet:

int i = 23;
int j = 1;
int base = 10;
int k = 2;
i += j * pow(base, k);
cout << i << endl;

Expectedly, this code should output "123," but instead, it surprisingly prints "122." This unexpected result may baffle you, especially if you're compiling with g 4.7.2 in a Windows XP environment.

At the heart of this discrepancy is the fact that std::pow() is designed to handle floating-point numbers. While these floating-point numbers offer convenience, they come with a trade-off: limited precision. In the case of std::pow(), the implementation may not handle exponentiation with perfect accuracy, leading to rounding errors.

To address this issue, one could harness the power of C 11 and define their own integer exponentiation function:

constexpr int int_pow(int b, int e) {
    return (e == 0) ? 1 : b * int_pow(b, e - 1);
}

This custom int_pow function is tailored specifically for integers, ensuring accurate results.

Alternatively, a tail-recursive approach proposed by Dan Nissenbaum is also a viable solution:

constexpr int int_pow(int b, int e, int res = 1) {
    return (e == 0) ? res : int_pow(b, e - 1, b * res);
}

With these options at hand, you can now confidently wield integer exponentiation in your C programs, banishing incorrect results to the realm of forgotten bugs.

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