algorithm - What is the most efficient to count trailing zeroes in an integer? -

normal way is

int main(){   int a=2000,count=0,temp;   while(a!=0)   {     temp=a%10;     if(temp==0) count++     else break;     a/=10;   }   printf("%d",count); } 

is there more efficient way ?

for 32-bit integer (maximum value 2147483647) need maximum of 4 tests. 64 bits add 1 more test 16 zeros.

start larger powers of 10 , work down:

int counttrailingzeros(int n) {     int zeros = 0;     if((n % 100000000) == 0)     {         zeros += 8;         n /= 100000000;     }     if((n % 10000) == 0)     {         zeros += 4;         n /= 10000;     }     if((n % 100) == 0)     {         zeros += 2;         n /= 100;     }     if((n % 10) == 0)     {         zeros++;     }     return zeros; } 

this has better worst-case performance, if 9/10ths of numbers pass have no trailing zeros average case worse. depends on values passing in typical case.

however, if 9/10ths of numbers pass have no trailing zeros shouldn't worry optimizing original code in first place since break in first iteration of loop 90% of time.