使用uint64将64位数字从二进制转换为十进制

本文介绍了使用uint64将64位数字从二进制转换为十进制的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我想将64位数字从二进制转换为十进制.由于dec2bin仅支持多达52位，因此我认为我可以滚动自己的函数并使用uint64来超出此限制:

I want to convert 64 bit numbers from binary to decimal. Since dec2bin only supports up to 52 bits, I thought I could roll my own function and use uint64 to go beyond this limit:

function [dec] = my_bin2dec(bin)
v = uint64(length(bin)-1:-1:0);
base = uint64(2).^v;
dec = uint64(sum(uint64(base.*(uint64(bin-'0')))));
end

但是，它不能按预期工作:

However, it does not work as expected:

my_bin2dec('111000000000000000000000000000000000001010110101011101000001110')

ans =

  8070450532270651392

my_bin2dec('111000000000000000000000000000000000001010110101011101000001111')

ans =

  8070450532270651392

这是正确的结果:

(111000000000000000000000000000000000001010110101011101000001110)bin
= (8070450532270651918)dec

(111000000000000000000000000000000000001010110101011101000001111)bin
= (8070450532270651919)dec

我想念什么?似乎仍有一些操作使用52位双精度算术执行，但我不知道是哪个.

What am I missing? It seems like there is some operation still performed using 52bit double arithmetic, but I don't know which one.

我检查了uint64是否可用的操作，并且似乎我使用的操作(power，times，sum)在那里:

I checked if the operations are available for uint64 and it seems that the ones I use (power, times, sum) are there:

>> methods uint64

Methods for class uint64:

abs         bitxor      diff        isinf       mod         plus        sum
accumarray  bsxfun      display     isnan       mpower      power       times
all         ceil        eq          issorted    mrdivide    prod        transpose
and         colon       find        ldivide     mtimes      rdivide     tril
any         conj        fix         le          ne          real        triu
bitand      ctranspose  floor       linsolve    nnz         rem         uminus
bitcmp      cummax      full        lt          nonzeros    reshape     uplus
bitget      cummin      ge          max         not         round       xor
bitor       cumprod     gt          min         nzmax       sign
bitset      cumsum      imag        minus       or          sort
bitshift    diag        isfinite    mldivide    permute     sortrowsc

推荐答案

您说的没错

问题出在

操作sum(uint64(base.*(uint64(bin-'0'))))给出double结果，该结果只有大约15个有效数字.这就是为什么您的最低位数是错误的.随后的转换为uint64并没有帮助，因为精度已经丢失.

The operation sum(uint64(base.*(uint64(bin-'0')))) gives a double result, which only has about 15 significant digits. That's why your lowest digits are wrong. Subsequent conversion into uint64 doesn't help, because precision has already been lost.

解决方案是将中的本机相加.这给出了具有完全精度的uint64结果:

The solution is to sum natively in uint64. This gives a uint64 result with its full precision:

dec = sum(uint64(base.*(uint64(bin-'0'))), 'native');

这篇关于使用uint64将64位数字从二进制转换为十进制的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

UInt64