长双精度浮点错误

Question

如果用C编写程序来了解重复除法的浮点误差的大小.

#include <stdio.h>

int main (int argc, char* argv[]) {
    if (argc < 3) {
        printf("Enter a decimal number as the first positional " 
                "argument\n");
        printf("Enter the maximum number of digits to print as the " 
                "second positional argument\n");
        return 0;
    }   

    long double d;
    sscanf(argv[1], "%Lf", &d);
    int m;
    sscanf(argv[2], "%d", &m);

    int i;
    char format[10];
    for (i = 1; i <= m; ++i) {
        printf("(%d digits)\n", i); 
        sprintf(format, "%%.%dLf\n\n", i); 
        printf(format, d); 
    }   

    long double p = d;
    printf("\n");
    for (i = 1; i <= m; ++i) {
        printf("(%Lf/10e%d with %d digits)\n", d, i, m); 
        p = p/(long double)10.0;
        printf(format, p); 
    }
    return 0;
}

使用以下参数运行时,这是输出的一行

$ fpe 0.1 700
.
.
.
(0.100000/10e180 with 700 digits)
0.0000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000999999999999999999969819570700939858153376
736698732853283605408116087882762948991724868957176649769045358705872354052
261113540314114885779914335315639806061208847920179776799404948795506248532
485303630811119507604985596684233990126219304092175565232198569923253737561
276484626462077772036038845251286782974821021132356946292172207615386395848
331484216638642723800290357587296443408362280895970909637712494349003491485
594533190659822910753768473307578901199121901299804449081420898437500000000
000000000000000000000000000
.
.
.

这里我们观察浮点噪声的485位数.这是用gcc 4.4.3编译的,我假设它是使用80位扩展精度.但是,485位十进制数字超过80位信息.所以,我的问题是,这些信息来自哪里？

Answer 1

没有打印额外信息.打印的值正是值p.

180次迭代后,p是+ 0x1.A8E90F9908E0CA56p-602,即15309010345804195115•2 ^-665.IEEE 754标准将浮点数的值定义为符号(+1或-1)乘以2的整数幂(由数字的指数字段确定)乘以其有效数的值(分数部分).所以每个浮点数都有一个特定的值.以上是价值p.在十进制,该值是完全.9999999999999999999698195707009398581533767366987328532836054081160878827629489917248689571766497690453587058723540522611135403141148857799143353156398060612088479201797767994049487955062485324853036308111195076049855966842339901262193040921755652321985699232537375612764846264620777720360388452512867829748210211323569462921722076153863958483314842166386427238002903575872964434083622808959709096377124943490034914855945331906598229107537684733075789011991219012998044490814208984375•10 ^-181.

这是您的计划产生的价值.因此,您的输出格式化程序已准确打印出值p.它做得很好.

事实上,在所有周围,浮点都做得很好.该值是最接近10 ^-181的长双^精度值.在长双程中不可能更接近.因此,即使经过数百次算术运算,错误也没有增加.

这里没有新的信息.如果我们被告知代表的位p,我们可以产生相同的数百个十进制数字.他们没有告诉你任何新的东西.但是,它们也不是垃圾; 它们完全由价值决定p.