is tha good method?

This is a discussion on is tha good method? within the C Programming forums, part of the General Programming Boards category; Heh Code: a = a + b; b = a - b; a = a - b; produces movl -8(%ebp), ...

  1. #16
    and the hat of wrongness Salem's Avatar
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    Heh
    Code:
        a = a + b;
        b = a - b;
        a = a - b;
    produces
    	movl	-8(%ebp), %edx
    	leal	-4(%ebp), %eax
    	addl	%edx, (%eax)
    	movl	-8(%ebp), %edx
    	movl	-4(%ebp), %eax
    	subl	%edx, %eax
    	movl	%eax, -8(%ebp)
    	movl	-8(%ebp), %edx
    	leal	-4(%ebp), %eax
    	subl	%edx, (%eax)
    vs.
    Code:
            int temp = a;
            a = b;
            b = temp;
    produces
    	movl	-4(%ebp), %eax
    	movl	%eax, -12(%ebp)
    	movl	-8(%ebp), %eax
    	movl	%eax, -4(%ebp)
    	movl	-12(%ebp), %eax
    	movl	%eax, -8(%ebp)
    10 instructions vs. 6
    and no undefined behaviour....


    > Temp variable swap takes more time because of the temp variable creation
    Creating 2 or 20 local variables takes the same amount of time, and since it's always done if you have any local variables at all (and takes only one instruction anyway), I fail to see your point.
    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.
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  2. #17
    Obsessed with C chrismiceli's Avatar
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    I thought that integer overflow was undefined behavior in ANSI?
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  3. #18
    & the hat of GPL slaying Thantos's Avatar
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    Unsigned integer under/overflow is defined behavior (it wraps around)
    signed interger under/overflow is undefined behavior

  4. #19
    Has a Masters in B.S.
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    its called C

    use it at your own risk... in other words dont use it unless you know what the hell your doing... this is whats wrong with modern languages... they assume your an idiot. This is why C will never die.

    Applies to everything... give up freedom/power for security/safty.... generally a bad idea most of the time...

    on the subject, is it a good method, depends on what your doing, but generally you should read what Salem said, let the optimizer do its job... cause it will do it better than you.
    Last edited by no-one; 07-25-2004 at 01:49 PM.
    ADVISORY: This users posts are rated CP-MA, for Mature Audiences only.

  5. #20
    Registered User linuxdude's Avatar
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    one "cute trick" that a book said was this
    Code:
    if(!var&1) /*number is even*/
    is that always defined? I know endianness is an issue for the MSB depending on which side it is on, so I don't know about this.

  6. #21
    & the hat of GPL slaying Thantos's Avatar
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    I'm not really sure on the order of precedence in that situation but there really is only two possibilities
    !(var&1) and (!var)&1

    So given var == 4 using eight bits:
    Code:
    var  == 00000100
    1    == 00000001
    !var == 11111011
    
    (var & 1)  == 00000000
    !(var & 1) == 11111111
    
    (!var & 1) == 00000001
    Well both come up with a true statement but lets check on a odd number (var == 3)
    Code:
    var  == 00000011
    1    == 00000001
    !var == 11111100
    
    (var & 1)  == 00000001
    !(var & 1) == 11111110
    
    (!var & 1) == 00000000
    Well the last one is the only one that came up with a false statement. Though as always I'd avoid bit-wise tricks whenever possible

  7. #22
    Registered User linuxdude's Avatar
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    well actually I just shortened it myself. ! is evaluated with & from left to right. The code in the book was
    Code:
    if((var&1)==0){

  8. #23
    & the hat of GPL slaying Thantos's Avatar
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    Heh I try not to memorize certain things, there is a tendacy to remember wrong at the worst possible times

  9. #24
    Registered User linuxdude's Avatar
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    don't worry I looked it up in C programming language[page 53 ]

  10. #25
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    Quote Originally Posted by linuxdude
    one "cute trick" that a book said was this
    Code:
    if(!var&1) /*number is even*/
    is that always defined? I know endianness is an issue for the MSB depending on which side it is on, so I don't know about this.
    As long as you don't access to the data byte by byte, there is no endianness issues. Bitwise operators are portable with unsigned integers (char, short, int, long...)
    Last edited by Emmanuel Delaha; 07-25-2004 at 03:00 PM. Reason: missing comma
    Emmanuel Delahaye

    "C is a sharp tool"

  11. #26
    and the hat of wrongness Salem's Avatar
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    > I know endianness is an issue for the MSB depending on which side it is on
    Doesn't matter what endian the machine is - big, little, middle or whatever
    var & 1 will always test the least significant bit
    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.
    I support http://www.ukip.org/ as the first necessary step to a free Europe.

  12. #27
    ATH0 quzah's Avatar
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    Quote Originally Posted by Thantos
    I'm not really sure on the order of precedence in that situation but there really is only two possibilities
    !(var&1) and (!var)&1

    So given var == 4 using eight bits:
    Code:
    var  == 00000100
    1    == 00000001
    !var == 11111011
    
    (var & 1)  == 00000000
    !(var & 1) == 11111111
    
    (!var & 1) == 00000001
    Well both come up with a true statement but lets check on a odd number (var == 3)
    Code:
    var  == 00000011
    1    == 00000001
    !var == 11111100
    
    (var & 1)  == 00000001
    !(var & 1) == 11111110
    
    (!var & 1) == 00000000
    Well the last one is the only one that came up with a false statement. Though as always I'd avoid bit-wise tricks whenever possible
    You've got a slight problem there... ! anything which is non-zero is zero. ! anything zero is 1.
    Code:
    #include <stdio.h>
    #include <limits.h>
    void showbits( unsigned int x )
    {
            int y;
            for( y = 0; y < sizeof x * CHAR_BIT; y++ )
                    putchar( (x & (1<<y)) ? '1' : '0' );
    }
    
    
    int main ( void )
    {
            unsigned int x = !4;
            int z;
    
            printf("x is %u\n", x );
            printf("Bits: ");
            showbits( x );
            printf("\n");
    
            return 0;
    }
    I don't know what you were meaning to show with your bit pattern, but it's wrong in the example where you're showing "!var". "!nonzero" is always zero. Which is why we can do instead:
    Code:
    void showbits( unsigned int x )
    {
            int y;
            for( y = 0; y < sizeof x * CHAR_BIT; y++ )
                    printf("%d", !!(x&(1<<y)) );
    }
    Unless I've misunderstood how you were trying to illustrate that.

    Quzah.
    Hope is the first step on the road to disappointment.

  13. #28
    & the hat of GPL slaying Thantos's Avatar
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    No your right quzah, for some reason I mixxed up ! and ~

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