c++:static_cast<int> std::sqrt(x) 是否总是为平方的正整数给出准确的结果？

Question

这个问题在这里已经有了答案:





C++ sqrt function precision for full squares

(3 个回答)


去年关闭。




是否保证，static_cast<int>(std::sqrt(x * x)) == x对于所有 x*x 不会溢出的正 x？
如果没有，我将如何稳健地计算这些数字的平方根？

Answer 1

来自 cppreference:

std::sqrt is required by the IEEE standard to be exact. The only other operations required to be exact are the arithmetic operators and the function std::fma. After rounding to the return type (using default rounding mode), the result of std::sqrt is indistinguishable from the infinitely precise result. In other words, the error is less than 0.5 ulp. Other functions, including std::pow, are not so constrained. (ref)

因此，鉴于您的限制，我看不出有任何担心的理由。