案例1 ：

for num in [.1,.2,.3,.4,.5,.6,.7,.8,.9,]:
    print(format(num,".50f"))

0.10000000000000000555111512312578270211815834045410
0.20000000000000001110223024625156540423631668090820
0.29999999999999998889776975374843459576368331909180
0.40000000000000002220446049250313080847263336181641
0.50000000000000000000000000000000000000000000000000
0.59999999999999997779553950749686919152736663818359
0.69999999999999995559107901499373838305473327636719
0.80000000000000004440892098500626161694526672363281
0.90000000000000002220446049250313080847263336181641

精度不高（.5除外）。

案例2 ：

for num in [1.,2.,3.,4.,5.,6.,7.,8.,9.]:
    print(format(num,".50f"))

1.00000000000000000000000000000000000000000000000000
2.00000000000000000000000000000000000000000000000000
3.00000000000000000000000000000000000000000000000000
4.00000000000000000000000000000000000000000000000000
5.00000000000000000000000000000000000000000000000000
6.00000000000000000000000000000000000000000000000000
7.00000000000000000000000000000000000000000000000000
8.00000000000000000000000000000000000000000000000000
9.00000000000000000000000000000000000000000000000000

完美的精度-???

众所周知，计算中没有完美的浮点整数：所有浮点都以二进制为基础来表示，精度取决于位大小（float32，float64等）。那么，上述案例2的处理方式是什么？即使对于".1000f"，零也将持续存在，这基本上意味着无限的精度。此外，0.5也以某种方式完美体现。

如果format不能强制Python打印浮点数的“ true”值，那怎么办？

尝试的替代方法：

1. format(round(num,50),".50f")
2. format(numpy.float128(num),".50f")
3. format(round(numpy.float128(num),".50f")
4. format("%.50f" % num)
5. "{:.50f}".format(num))
6. f"{num:.50f}"

接受的答案：阐明问题中假定的错误前提；实际问题的答案就在问题本身之内-使用format显示真实的数值。