The rookie series of the C + + Classic questions (eight)

Source: Internet
Author: User

calculating binary in binary 1 the number

topic: bit arithmetic programming is seldom encountered, but it's also a heavy one, just a point, a more common topic is to calculate the binary representation of a number. 1 the number of.

Analysis: A 1 is not 0, our idea is a one of the operation, we quickly think of the following practices:

int countBit1 (int val) {    Register int count = 0;    while (Val)    {        if (1 & val)        {            ++count;        }        Val >>= 1;    }    return count;}

A look at this implementation is good, look carefully there will be a big problem, that is, when val is negative, there will be a dead loop, Because the sign is the arithmetic shift , we will take into account the problem of the sign bit, so we have modified this, in exchange for a way of thinking:

int countBit2 (int val) {    Register int count = 0;    Register unsigned int flag = 1;    while (flag)    {        if (Val & flag)        {            ++count;        }        Flag <<= 1;    }    return count;}

Well, this implementation solves one of the above unsigned problems, but there is a drawback, that is , regardless of the number of 1 of the number of a few, have to cycle three times (sizeof (int) = the time ), This does not meet the efficiency requirements, so take a closer look at the implementation of one, the problem is that there is now a symbolic shift, we look at the strong system to convert signed to unsigned in the operation :

int countBit3 (int val) {    Register int count = 0;    Register unsigned int _val = val;    while (_val)    {        if (1 & _val)        {            ++count;        }        _val >>= 1;    }    return count;}

Well, good work, this implementation is OK, but for the pursuit of the perfect programmer, can not stop, we can have a few 1 on the loop several times, and, the circulation body has branch jump, which will affect performance, so I think:

For example: x = 1101 Number, we perform the following operations:

x–1 = 1100, the attentive person will find that when the lowest bit is 1, only change the most, when the bit is 0, the lowest bit of 1 and the left bit are reversed, so we can use N &= (n–1) to clear the right 1.Whyn &= (n–1)to clear the far right.1it? Because from the point of view of the binary,Nequivalent ton-1the lowest bit plus1. To give an example,8( +)= 7(0111)+ 1(0001), so8 & 7 =( +)&(0111)= 0(0000), clear the8on the far right.1(in fact, the most high1, because8In binary , there is only one1). Another example7(0111)= 6(0110)+ 1(0001), so7 & 6 =(0111)&(0110)= 6(0110), clear the7the binary representation of the rightmost1(i.e. the lowest bit1)

The implementation is as follows:

int countBit4 (int val) {    Register int count = 0;    while (Val)    {        ++count;        Val &= val-1;    }    return count;}
Of course there are other ways to do it, and I'm not going to list it here.

If there is a mistake welcome to point out, share please identify

Feel good words on the top one, feel good words on the step.

The rookie series of the C + + Classic questions (eight)

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