Strlen

[_arjun]

If I write some thing like this

for(int i = 0 ; i < strlen(s) ; i++)

will it be O(n^2) ?

Also please tell me where can I see codes of standard library of C (GCC) in Ubuntu.

[grumpy]

The complexity depends what the loop body does. It is, however, not a good idea to evaluate strlen(s) every iteration if you care about efficiency, unless the contents of s changes within the loop body. Try evaluating strlen(s) before the loop, once only.

You can download the source for various versions of gcc from somewhere on this site. Somewhere among those source distribution will be the source for libc (if I remember the name correctly), which is gnu's corresponding implementation of the C standard library.

[memcpy]

I'm not exactly sure, but isn't your for loop actually O(n*i)?

[_arjun]

Thanks

[megafiddle]

Wouldn't it just be directly proportional to the string length (linear)?

O(n)?

[christop]

strlen(s) has O(n) run time, where n is the length of the string. strlen(s) is called n times, so yes, the loop is O(n^2). As grumpy said, you should call strlen(s) once before the loop.

EDIT: strlen could theoretically be implemented with O(1) run time, such as on a processor that can find the first 0 byte in constant time, but this is only theoretical. Assume strlen is O(n).

[megafiddle]

Of course, missed that.

Basically an O(n) nested in an O(n).

[iMalc]

The reason "The complexity depends what the loop body does." is that the loop might say append characters onto the string s, or put a null-terminator earlier in the string, or it might have another loop inside this one etc.

Without anything special going on, it is as was originally guessed.

[Malcolm McLean]

If I write some thing like this

for(int i = 0 ; i < strlen(s) ; i++)

will it be O(n^2) ?

Yes. strlen steps through the sgtring character by character to find the nul. Your callin it N times. So that's N squared operations.

[_arjun]

Thank you all guys