How to solve the problem of bit-by-bit extraction and binary conversion about C++ read-in
Today, I'm going to talk to you about how to solve the problem of bit-by-bit extraction and binary conversion of C++ readings. maybe many people don't know much about it. in order to make you understand better, the editor summed up the following. I hope you can get something from this article.
I will not write a specific question on this blog, but just sum up a typical problem-reading in numbers and taking them out bit by bit.
Take the number 12345, for example.
Yes, first of all, we need to take out a position. Take it out like this:
12345 take 12345
12345mm 5. / / in order to find the rules
So we have its position. Ten of you are like this:
12345 take 10 times 1234
1234404.
By the same token, 100:
12345ax 100mm 123
123 to 3.
So you can find out which one to take out is to divide the primitive number by the bit name of this bit, and then modulo 10.
Program:
# include#includeusing namespace std;int main () {int a [100]; int wei = 0; int num; cin > > num; while ((num/ (int) pow)! = 0) / the termination condition of the loop is that the number of digits of this number is less than the number of digits to be divided this time {a [wei] = (num/ (int) pow (10Magnewei)); / / according to the conclusion just reached, take out all of you and save them in the array. Wei++;}}
Then there is the problem of binary conversion. In fact, it is similar to the bit problem, except that after it is taken out, it has to be multiplied by the power of the corresponding bit.
Program:
Long long to10 (int jz,int num) / / function: convert the input number to decimal {long long result=0; int wei=0; while (num/ (int) pow (10jiawei)! = 0) / / take out the input number bit by bit {result+=pow (jz,wei) * ((int) (num/pow (10wei); / / multiply the position of the number by the corresponding power wei++ of the decimal system / / (num/1) / / (num/10) / / (num/100)} return result;}
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