Q:
C program to Count the Number of Trailing Zeroes in an Integer
belongs to collection: C solved programs/examples on Bitwise Operators
C solved programs/examples on Bitwise Operators
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C programming
Pre-requisite: Input number n
Algorithm:
1) Set count=0 2) Do bit wise AND with n and 1. n& 1 let n be a7a6a5a4a3a2a1a0 1->00000001 So doing bitwise AND (refer to published article on bitwise operators) will result in all bits 0 except the LSB which will be a0. If a0 is 0 then the Integer has trailing zero else don’t have Thus, IF n& 1 = =0 count++ ELSE Break; //no more trailing zero since current LSB is 1 END IF-ELSE 3) Right shift n by 1 n=n>>1 4) IF n==0 Break ELSE REPEAT step 2, 3 5) PrintExample with Explanation:
Trailing zeroes in 12: 12 → 00001100
So first iteration: n=12 //00001100 n & 1 =0 count=1 n=n>>1 (n=6) //00000110 So second iteration: n=6 //00000110 n & 1 =0 count=2 n=n>>1 (n=3) //00000011 So third iteration: n=3 //00000011 n & 1 =1 break print count=2 No of trailing zeroes: 2C implementation
Output (first run)
Output (second run)
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