本文整理匯總了Golang中cmd/compile/internal/big.Float.Mul方法的典型用法代碼示例。如果您正苦於以下問題：Golang Float.Mul方法的具體用法？Golang Float.Mul怎麽用？Golang Float.Mul使用的例子？那麽, 這裏精選的方法代碼示例或許可以為您提供幫助。您也可以進一步了解該方法所在類cmd/compile/internal/big.Float的用法示例。
在下文中一共展示了Float.Mul方法的1個代碼示例，這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚，您的評價將有助於係統推薦出更棒的Golang代碼示例。
示例1: Example_sqrt2
// This example shows how to use big.Float to compute the square root of 2 with
// a precision of 200 bits, and how to print the result as a decimal number.
func Example_sqrt2() {
// We'll do computations with 200 bits of precision in the mantissa.
const prec = 200
// Compute the square root of 2 using Newton's Method. We start with
// an initial estimate for sqrt(2), and then iterate:
// x_{n+1} = 1/2 * ( x_n + (2.0 / x_n) )
// Since Newton's Method doubles the number of correct digits at each
// iteration, we need at least log_2(prec) steps.
steps := int(math.Log2(prec))
// Initialize values we need for the computation.
two := new(big.Float).SetPrec(prec).SetInt64(2)
half := new(big.Float).SetPrec(prec).SetFloat64(0.5)
// Use 1 as the initial estimate.
x := new(big.Float).SetPrec(prec).SetInt64(1)
// We use t as a temporary variable. There's no need to set its precision
// since big.Float values with unset (== 0) precision automatically assume
// the largest precision of the arguments when used as the result (receiver)
// of a big.Float operation.
t := new(big.Float)
// Iterate.
for i := 0; i <= steps; i++ {
t.Quo(two, x) // t = 2.0 / x_n
t.Add(x, t) // t = x_n + (2.0 / x_n)
x.Mul(half, t) // x_{n+1} = 0.5 * t
}
// We can use the usual fmt.Printf verbs since big.Float implements fmt.Formatter
fmt.Printf("sqrt(2) = %.50f\n", x)
// Print the error between 2 and x*x.
t.Mul(x, x) // t = x*x
fmt.Printf("error = %e\n", t.Sub(two, t))
// Output:
// sqrt(2) = 1.41421356237309504880168872420969807856967187537695
// error = 0.000000e+00
}