Thread: Which string operation is faster? Which is better?

  1. #1
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    Which string operation is faster? Which is better?

    If I have a string passed in that has some characters at the beginning of it that each signify a type of info and I want to put it in a loop that each time strips off one more character from the beginning, does it make more sense to use "substr" or the pointer?

    Code:
    //Example one
    void doSmth(std::string &str, int length)
    {
         for ( int i = 0; i < length; i++ )
         {
              //some code elided here
              ...
              else doSmth( str.substr( i ), length - i );
         }
    }
    
    //Example two
    void doSmth(std::string &str, int length)
    {
         for ( int i = 0; i < length; i++ )
         {
              //some code elided here
              ...
              else doSmth( *(++str), length - i );
         }
    }

  2. #2
    Officially An Architect brewbuck's Avatar
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    Calling substr() in a loop would be very inefficient. Just loop through the string counting the prefix characters, until you reach the end of the prefix. Then make a single call to substr() to extract the prefix (or the remainder, whatever the case may be).

  3. #3
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    Quote Originally Posted by brewbuck View Post
    Calling substr() in a loop would be very inefficient. Just loop through the string counting the prefix characters, until you reach the end of the prefix. Then make a single call to substr() to extract the prefix (or the remainder, whatever the case may be).
    What about *(str++)?

  4. #4
    Officially An Architect brewbuck's Avatar
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    Quote Originally Posted by 6tr6tr View Post
    What about *(str++)?
    Since str++ is illegal, I would have to say that option is bad.

  5. #5
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    Is length the length of the string? If so, you don't need to pass that around, the string class knows its length (call length() or size()).
    Code:
    void doSmth(std::string &str, int prefixlength)
    You could pass the prefix length if you wanted to implement this recursively, or you could do it iteratively and just remember it.

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