Thread: Requesting help on problem regarding using "^=" on strings

  1. #1
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    Requesting help on problem regarding using "^=" on strings

    Hi all, I am having trouble understanding the answer for this exercise in KNKING c programming a modern approach book

    make a function
    (d) strrev(s) -- Reverses the characters in s (except the null character); returns s.

    Code:
    char *strrev(char *s)
    {
        char *t = s, *p;
        p = s + strlen(s) - 1;
        while (t < p)
        {
            *t ^= *p;
            *p ^= *t;
            *t ^= *p;
            t++;
            p--;
        }
        return s; }
    
    I understand t is a pointer to the first character in the string, and p is the last pointer. I know that ^ is a bitwise exclusive OR operator, but i have no clue what the first 3 lines inside the while loop do, and how they swap the character around. Everything else I understand.

    Thank you all for any help that can be provided. Cheers

  2. #2
    and the hat of int overfl Salem's Avatar
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    Question 20.15c

    Use a temporary variable in your code.
    It's a lot clearer to basically anybody what your intention is.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.

  3. #3
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    XOR has 3 basic uses: To invert bits based in a mask; To get the result of different bits in both operands or as a caryless add.
    This kind of code is a trick to swap values without using a temporary variable, but it isn't a good one. Take a look at this two optimized (by the compiler, using -O2 option) functions f() and g():
    Code:
    ; void f( int *a, int *b )
    ; { int t; t = *a; *a = *b; *b = t; }
    f:
      mov eax,[rdi]
      mov edx,[rsi]
      mov [rdi],edx
      mov [rsi],eax
      ret
    
    ; void g( int *a, int *b )
    ; { *a ^= *b; *b ^= *a; *a ^= *b; }
    g:
      mov eax,[rdi]
      xor eax,[rsi]
      mov [rdi],eax
      xor eax,[rsi]
      mov [rsi],eax
      xor [rdi],eax
      ret

  4. #4
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    Thank you for the responses. Cheers a lot !

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