长双的用法

函数的目的是用牛顿 - 拉夫森法计算一个数的平方根。 我在while循环中包含了printf例程,所以我可以看到 root 2的值越来越接近实际值。 我最初使用float来定义epsilon,但是当我增加了epsilon的值时,返回结果的值似乎在一定数量的数字后被截断。 所以我决定把所有的变量都切换成长整倍数,程序显示的是负值结果。 我如何解决它?

//Function to calculate the absolute value of a number

#include <stdio.h>

long double absoluteValue (long double x)
{
    if (x < 0)
        x = -x;
    return (x);
}

//Function to compute the square root of a number

long double squareRoot (long double x, long double a)
{
    long double guess = 1.0;

    while ( absoluteValue (guess * guess - x) >= a){
        guess = (x / guess + guess) / 2.0;
        printf ("%Lfn ", guess);
        }

    return guess;
}

    int main (void)
    {
    long double epsilon = 0.0000000000000001;

    printf ("nsquareRoot (2.0) = %Lfnnnn", squareRoot (2.0, epsilon));
    printf ("nsquareRoot (144.0) = %Lfnnnn", squareRoot (144.0, epsilon));
    printf ("nsquareRoot (17.5) = %Lfn", squareRoot (17.5, epsilon));

    return 0;
}

如果你正在使用mingw的Code :: Blocks版本,请看这个答案:C中long double的转换说明符

mingw ... printf does not support the 'long double' type.

一些更多的支持文档。

http://bytes.com/topic/c/answers/135253-printing-long-double-type-via-printf-mingw-g-3-2-3-a

如果你直接从float变为long double ,那么你可以尝试使用double ,这是浮动状态的两倍,并且你可能不需要一路走到long double

为此,您可以使用%lf的打印说明符,并且您的循环可能想要看起来像这样,以防止基于您的epsilon的无限循环:

double squareRoot (    double x,     double a)
{
    double nextGuess = 1.0;
    double lastGuess = 0.0;

    while ( absoluteValue (nextGuess * nextGuess - x) >= a && nextGuess != lastGuess){
        lastGuess = nextGuess;
        nextGuess = (x / lastGuess + lastGuess) / 2.0;
        printf ("%lfn ", nextGuess);
        }

    return nextGuess;
}
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