如何获取整数的最后几位 [英] How to get the last bits of integer

问题描述

大家好，



假设我有以下计划

Hi All,

suppose i have the following program

main()
{
int a;
printf("enter the value of a\n");
scanf("%d\n",&a);

}

现在，如果我输入数字2，它将被存储为00000010 in a我想得到最后两位（32位系统）的唯一。

如何获得。

Now if i enter number 2 it will be stored as 00000010 in a.I want to get the last two bits(32-bit system) of a only.
How to get that.

推荐答案

试试：

try:

a = a & 3;

您可能需要了解位操作和二进制数。

你可以先用纸和笔进行实验：

- 什么是二进制代表说0,1,2,3,4 ......，15？

- 以下运算符是什么：bit-and（&），bit-or（|），bit-xor（^），complement（〜）？

使用4位价值仅限于开始。



例如十进制3 =二进制0011.



最右边的二进制数字是因子1，左边的下一个数字是2，然后是4然后是8。

这类似于十进制数：最右边是因子1，然后是10，然后是100，等等。

十进制：1 = 10 ⁰ ，10 = 10 ¹ ，100 = 10 ² 等。

二进制：1 = 2 ⁰ ，2 = 2 < sup> 1 ，4 = 2 ² 等。



Eg十进制123是1x10 ² + 2x10 ¹ + 3x10 ⁰ = 100 + 20 + 3

例如二元1110是1x2 ³ + 1x2 ² + 1x2 ¹ + 0x2 ⁰ =十进制8 + 4 + 2 + 0 = 14



对于位操作：操作是按位的：

You might need to learn about bit manipulations and binary numbers.
You may experiment first with paper and pencil:
- what is ther binary represntation of say 0, 1, 2, 3, 4, ..., 15?
- what do the the following operators: bit-and (&), bit-or (|), bit-xor (^), complement (~)?
Work with 4 bit values only to start with.

E.g. decimal 3 = binary 0011.

The right most binary digit has the factor 1, the next on the left is 2, then 4 and then 8.
This is analogous to decimal numbers: the right most is factor 1, then 10, then 100, etc.
Decimal: 1 = 10⁰, 10 = 10¹, 100 = 10², etc.
Binary: 1 = 2⁰, 2 = 2¹, 4 = 2², etc.

E.g. decimal 123 is 1x10² + 2x10¹ + 3x10⁰ = 100 + 20 + 3
E.g. binary 1110 is 1x2³ + 1x2² + 1x2¹ + 0x2⁰ = decimal 8+4+2+0 = 14

For bit-operations: the operations are bitwise:

x & y: result bit is 1 if x==1 and y==1, 0 otherwise    (bit-and)
x | y: result bit is 1 if x==1 or  y==1, 0 otherwise    (bit-or)
x ^ y: result bit is 1 if x!=y,          0 otherwise    (bit-xor)
~x:    result bit is 1 if x==0,          0 otherwise    (complement)

Eg十进制6& 3 =二进制0110& 0011 = 0010.

您可能更喜欢这样写：

E.g. decimal 6 & 3 = binary 0110 & 0011 = 0010.
You might prefer to write it this way:

6 = 0110
3 = 0011
& = 0010   the bits that are both 1 result in 1, all others in 0

所以，你的问题是只得到最低的两位：

So, your question on getting the lowest two bits only:

int a;
...
a = a & 3; // 3 = binary ...0011
...

这导致所有位，但最低的两位始终为0，最低的两位按原样。



干杯

Andi

This results in all bits but the lowest two are always 0, and the lowest two bits are taken as-is.

Cheers
Andi

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