Dark Horse Programmer---C language---memory profiling

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Bitwise AND &

1101110 101101

0110101 000100

——————— ---------------

0100100 000100

in situ and 1&, remained unchanged; and 0&, all 0

Therefore, you can let the original number of a bit with 1&, according to the result can determine whether the bit is 0 or 1, and the result is only 0 or 1

A&1==1//a is odd because the last digit of the odd number is 1.

A&1==0//a is even, because the last digit of the even number is 0.

outputs a binary form of an integer

void printbinary (int num) {

// with the right shift, each bit of &1 Gets the value of each bit printed

// define variables, record the number of bits that are shifted right

// right Shift after &1 bit gets the first bit value 0 or 1

//(sizeof (num) <<3) -1==4*8-1==31

for ( int count= (sizeof) <<3)-1; count>=0; count-- ) {

printf ("%d",(num>>count) &1); any number &1 results can only be 0 or 1

If (! ( count%4)) {

printf ("");

If (! ( count%8)) {

printf ("");

}

}

}

printf ("\ n");

}

Bitwise OR |

010110 101101

011100 000000

——————— ---------------

011110 101101

A|0=a

Bitwise XOR or ^

Each one is 0 and the difference is 1. and can be exchanged

A^b^c==a^c^b==b^a^c

A^a==0

A^0==a

A^b^a==b

The interchange A^b^a==a^a^b==0^b==b of A and B values using bitwise XOR or ^ operations

A=a^b;

B=a^b;

A=a^b;

Shift left <<

a<<n==a*2 N-Th square

But left shifts can lead to positive and negative changes.

Shift Right >>

A>>N==A/2 N-Th square

Positive and negative changes

Dark Horse Programmer---C language---memory profiling