PI和浮点数的准确性

Question

Pi的单/双/扩展精度浮点表示精确到小数位数？

Answer 1

#include <stdio.h>

#define E_PI 3.1415926535897932384626433832795028841971693993751058209749445923078164062

int main(int argc, char** argv)
{
    long double pild = E_PI;
    double pid = pild;
    float pif = pid;
    printf("%s\n%1.80f\n%1.80f\n%1.80Lf\n",
    "3.14159265358979323846264338327950288419716939937510582097494459230781640628620899",
    pif, pid, pild);
    return 0;
}

结果:

[quassnoi #] gcc --version
gcc (GCC) 4.3.2 20081105 (Red Hat 4.3.2-7)

[quassnoi #] ./test

3.14159265358979323846264338327950288419716939937510582097494459230781640628620899

3.14159274101257324218750000000000000000000000000000000000000000000000000000000000
        ^
3.14159265358979311599796346854418516159057617187500000000000000000000000000000000
                 ^
3.14159265358979311599796346854418516159057617187500000000000000000000000000000000
                 ^
  0000000001111111
  1234567890123456

Answer 2

当我检查Quassnoi的答案似乎可疑,我认为long double并double让我在一个小挖最终会得到相同的精度.如果我运行他用clang编译的代码,我得到了与他相同的结果.但是我发现如果我指定long double后缀并使用文字初始化long double,它提供了更多的精度.这是我的代码版本:

#include <stdio.h>

int main(int argc, char** argv)
{
    long double pild = 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899L;
    double pid = pild;
    float pif = pid;
    printf("%s\n%1.80f\n%1.80f\n%1.80Lf\n",
        "3.14159265358979323846264338327950288419716939937510582097494459230781640628620899",
        pif, pid, pild);
    return 0;
}

结果如下:

3.14159265358979323846264338327950288419716939937510582097494459230781640628620899

3.14159274101257324218750000000000000000000000000000000000000000000000000000000000
        ^
3.14159265358979311599796346854418516159057617187500000000000000000000000000000000
                 ^
3.14159265358979323851280895940618620443274267017841339111328125000000000000000000
                    ^

Answer 3

6位和14位。3位中的1位超过0，最后一位虽然已存储，但不能被视为精度点。

抱歉，如果没有更多上下文，我不知道扩展意味着什么。你是说C#的小数吗？