The C language converts integers in reverse in binary order

The source of the problem, this morning and a roommate to eat breakfast when talking about a problem, an integer in reverse order of binary, and then output the value after the reverse.

We know that the values in memory are stored in binary form, if we are 32-bit machine, every 8 bits is a byte, int type on the 32-bit machine is accounted for 4 bytes, that is 32 bits.

such as 2 = 0000 0000 0000 0000 0000 0000 00000 0010 (32-bit)

Inverse ^2 = 0100 0000 0000 0000 0000 0000 00000 0000 (here with ^ for reversal)

So how does this work? First, add some knowledge:

1) A = a << 1, means to move a to the left one bit, such as: 0010->0100 (generally back is 0)

2) b = b >> 1, means to move B to the right one bit, such as: 0100->0010 (generally preceded by a complement of 0)

3) B & 1, this means bitwise AND operation, such as: 2 & 1, in fact, the following actions are performed:

0000 0000 0000 0000 0000 0000 00000 0010 = 2

0000 0000 0000 0000 0000 0000 00000 0001 = 1

0000 0000 0000 0000 0000 0000 00000 0000 = 2 & 1 = 0

This operation, the first 31 bits are all set to 0, only the last one remains unchanged, the effect is to remove the last value.

4) A &=, does this need to be explained? Same as 3), but ~ indicates the position of the 0, set to 1;1 position, set to 0.

5) A |= 1, which represents a bitwise OR operation (A = a | 1), for example: 2 | 1, the following actions are actually performed:

0000 0000 0000 0000 0000 0000 00000 0010 = 2

0000 0000 0000 0000 0000 0000 00000 0001 = 1

0000 0000 0000 0000 0000 0000 00000 0011 = 2 | 1 = 3

OK, here's a look at the following code .....

Current Environment: win7_32bit,vs2010,c++

1#include <stdio.h>2 3 intMainvoid)4 {5     inti = +, a =2;//32-bit 0000 0000 0000 0000 0000 0000 0000 0010 = 26     intb = A;//Save a copy7 8      while(i--)9     {TenA = a <<1; OneA &= ~1;//to = 1111 1111 1111 1111 1111 1111 1111 1110 Make sure the 31st bit is 0 A         if(B &1)//1 = 0000 0000 0000 0000 0000 0000 0000 0001 -         { -A |=1;//Make sure the 31st bit is 1. the         } -b = b >>1; -     } -  +printf"%d\n", a); -  +     return 0; A}

Thought: The general idea is:

1) First make the value of a, B equal;

2) Then, one at a time from the tail of B (from the 32nd position to the No. 0 position, with I loop control); Note: b = b >> 1,b has been shifted right to ensure that the last one is removed each time.

3) Finally, append it to the end of a. Note: a = a << 1,a has been shifted to the left to ensure that after 32 cycles, the first appended number is reached at the end of the first digit.

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