c# - Binary Search with Strings -

i have object array , want search date / day property (user decides) using binary search.

**edit: read sharesarray[mid].date values (and day values) text file , examples are: 05/02/2015, 14/10/2014.the searchstring value gained user in same format date values. **

this first attempt trying date property:

int high, low, mid; high = sharesarray.length - 1; low = 0;  while (low <= high) {     mid = (low + high) / 2;     if (string.compare(sharesarray[mid].date, searchstring, true) == 0)     {         break;     }     else if (string.compare(sharesarray[mid].date, searchstring, true) > 0)     {         high = mid - 1;     }     else if (string.compare(sharesarray[mid].date, searchstring, true) < 0)     {         low = mid + 1;     } 

i have tried last else if statement else , doesn't make difference.

also, matter way round string1 , string2 in string.compare part?

summarizing answer q&a in comments of question, bit of additional context.

for binary search, there few ingredients need take care of

1. the input array binary search function must sorted.
2. given algorithm have implemented, must sorted in ascending order.
3. you must have three-way comparison function, compare(lhs, rhs) returns: < 0 if lhs < rhs, > 0 if lhs > rhs , 0 if lhs == rhs.
4. the order of arguments in compare function matters, if switch them, taking wrong branch , changing upper search bound instead of lower, or other way around.
5. for implementation, had order correct, lhs : sharesarray[mid].date , rhs = searchstring.
6. because comparing dates, need use datetime.compare function, , convert date strings datetime instances, using datetime.parseexact(mystring, "dd/mm/yyyy", cultureinfo.invariantculture);. string comparison wrong results such "05/02/2015" < "14/10/2014".

and there 1 more anecdotal remark make (see here):

the calculation of pivot value mid susceptible integer overflow. if calculate

mid = (low + high) / 2; 

then low + high become larger fits in integer.

the preferred way of calculating mid therefore

`mid = low + ((high - low) >> 1);`.