请解释 [英] Plz explain

问题描述

int多项式& = 0xFFFF;

请解释这行我没听懂.
在广告中表示感谢.

int polynom &= 0xFFFF;

Plz explain this line I am not geting.
thanks in adv.

推荐答案

它是一行垃圾.

Its a line of rubbish.

int polynom

声明一个称为"polynom"的整数并将其值设置为零.

Declares an integer called "polynom" and sets it''s value to zero.

polynom &= 0xFFFF

在多项式"的当前值和十六进制FFFF(十六位，全为"1")之间执行逻辑与，并将结果替换为多项式"的当前值.
即

Performs a logical AND between the current value of "polynom" and hexadecimal FFFF (sixteen bits, all ''1''s), and replaces the current value of "polynom" with the result.
I.e.

int polynom &= 0xFFFF;

声明一个称为"polynom"的整数并将其值设置为零-因为从逻辑上讲，0与任何值进行逻辑与运算就得出了零.

declares in integer called "polynom" and sets it''s value to zero - since 0 logically ANDed with anything gives zero by definition.

删除&

这将把polynnom声明为一个int并将其值设置为0xFFFF(十进制数为35535)

Remove the &

This will declare polynnom as an int and set its value to 0xFFFF (65535 decimal)

如果多项式是16位数字，则按位进行运算，其值为16 1"，因此不会有发生
如果多项式> 16位，则它将删除> 2 ^ 16

If the polynom is a 16 bit number then you bitwise and it with 16 1''s hence nothing will happen
if polynom > 16 bit then it will remove the bits >2 ^16

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