Why can you have mantissa effect?
为什么会有尾数效应？
The mantissa is basically a 23-bit unsigned integer - simple enough.
尾数一般是一个23位的不带符号的整数—它非常简单。
Wait a minute, didn't I say that the mantissa ranged from 0 to 33,554,431?
等一下，我刚才是不是说尾数的范围是从0 到 33,554,431 ？
Putting these together, a float is interpreted as sign * mantissa * 2exponent.
综合起来，浮点数是这样表示的：sign*mantissa*2exponent 。
But a look at passing motor vehicles, basic no mantissa for grades 1 and 6 vehicles.
但细看驶过的一辆辆机动车，基本没有牌号尾数为1和6的车辆。
The mantissa can also be found through bit masking, though the algorithm is far less obvious.
尽管算法不是很明显，但还是可以通过位屏蔽来查找尾数。
This is a standard floating point representation that the first bit of mantissa must be 1.
这是一个标准的浮点表示法，特点是尾数的第一个元不可以是零。
Of course, both exponent and mantissa are encoded using a binary system rather than decimal system.
当然，幂和尾数都是用二进制数系统来编码，而不是十进制数系统。
Obviously with finite precision the repetition in the mantissa will be cut off in different places.
显然，有限精度尾数的重复会在不同的地方切断。
Of course, both exponent and mantissa are encoded using a binary system rather than a decimal system.
当然，幂和尾数都是用二进制数系统来编码，而不是十进制数系统。
The final 23 bits hold the mantissa (sometimes called the significand), which ranges from 0 to 33,554,431.
最后的23位表示尾数（有时称为有效数字），其值的范围是0到 33,554,431。
In the language of computer science precision is given in terms of the size of the mantissa and exponent.
在计算机科学的精确语言的尾数和指数的大小了。
Doubles are encoded in much the same way except that they use a 52-bit mantissa and an 11-bit exponent for higher precision.
双精度数是以类似的方式编码的，但是它使用52位的尾数和11位的指数来获得更高的精度。
In the case of the similar stocking density (mantissa), the large-sized fish played a distinct role in increasing capital gains rate of pond productivity.
在放养密度(尾数)近似的情况下，大规格鱼种对增加池塘生产力资金增益率有显著作用。
In a numeration system, the number that is raised to the power denoted by the exponent and then multiplied the mantissa to determine the number represented.
在固定基数记数制中，一种通过对给定数进行运算而得到的数。
A problem with adjusting the exponent of one operand so that it matches the exponent of the other operand is that we only have so many bits to use to represent the mantissa.
调整某个操作数的指数以使其匹配其他操作数指数的一个问题是，我们只有同样多的位数可用以表示尾数。
A problem with adjusting the exponent of one operand so that it matches the exponent of the other operand is that we only have so many bits to use to represent the mantissa.
调整某个操作数的指数以使其匹配其他操作数指数的一个问题是，我们只有同样多的位数可用以表示尾数。
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