Re: reverse a string with 0 terminator in one pass

From:
pjb@informatimago.com (Pascal J. Bourguignon)
Newsgroups:
comp.lang.c++
Date:
Wed, 04 Mar 2009 14:59:01 +0100
Message-ID:
<7ck575qti2.fsf@pbourguignon.anevia.com>
James Kanze <james.kanze@gmail.com> writes:

On Mar 4, 12:40 am, Juha Nieminen <nos...@thanks.invalid> wrote:

dbtouch wrote:

I have been thinking solution for this problem: to reverse a string
with 0 terminator in one pass without using auxillary memory. My
initial thought is to use recursion, but I have not succeeded finding
an answer.


I don't even think the problem is solvable strictly in one
single pass even if you are allowed to use additional memory
freely.


Sure it is:

    void
    reverse( char* source )
    {
        std::size_t cap = 32 ;
        char* result
            = static_cast< char* >( std::malloc( cap + 1 ) ) ;
        assert( result != NULL ) ;
        *result = 0 ;
        for ( std::size_t i = 0 ; source[ i ] != '\0' ; ++ i ) {
            if ( cap <= i ) {
                cap += cap / 2 ;
                result = static_cast< char* >(
                                realloc( result, cap + 1 ) ) ;
                assert( result != NULL ) ;
            }
            std::memmove( result + 1, result, i + 1 ) ;
            result[ 0 ] = source[ i ] ;
        }
        std::strcpy( result, source ) ;
        std::free( result ) ;
    }


You are cheating. It all depends on the definition of pass.

We could agree that one pass is accessing each elements of a vector
once (let's say one read and one write to each element of a vector).

The stack is itself to be considered a vector, pushing is writting,
poping is reading. Hiding a pass inside a function doesn't remove it.

In your reverse you have three passes at least:
- malloc may initialize the allocated memory block, 1/2 pass,
- for loop, 1/2 pass (just reading source),
- memmove, 1 pass,
- strcpy, 1 pass.

The most efficient way to implement it is to scan for the null byte
(1/2 pass) and to do the reversing (1 pass), for a total of 1.5
"passes".

        std::string tmp( source ) ; 1 pass
        std::reverse( tmp.begin(), tmp.end() ) ; 1 pass
        std::strcpy( source, tmp.c_str() ) ; 1 pass

-----------------------------------------------------------

Which is probably more efficient than my code, above, but I'm
willing to bet that there's a call to strlen hidden somewhere in
the constructor of std::string that I use.


You doubt it?

The whole question is, of course, silly.


Indeed.

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__Pascal Bourguignon__

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