Why does the value get discarded in this case?

Discussion in 'C Programming' started by Chad, Jun 6, 2007.

  1. Chad

    Chad Guest

    When I have:

    int main(void)
    int x = 256;

    printf("The value is: %d\n", x);
    return 0;

    I get:

    [cdalten@localhost ~]$ gcc -g -Wall seq.c -o seq
    seq.c: In function 'main':
    seq.c:6: warning: statement with no effect
    [cdalten@localhost ~]$ ./seq
    The value is: 256

    However, when I change x from x>>8 to x++

    #include <stdio.h>

    int main(void)
    int x = 256;

    printf("The value is: %d\n", x);
    return 0;

    I get:

    [cdalten@localhost ~]$ gcc -g -Wall seq.c -o seq
    [cdalten@localhost ~]$ ./seq
    The value is: 257

    The question is, how come something like x>>8 discards the value right
    away, but x++ doesn't?
    Chad, Jun 6, 2007
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  2. Chad said:

    It does. The difference is that x++ has a side-effect, which x>>8
    doesn't, and the compiler considers it plausible that you wrote x++ not
    for its value but for its side-effect, so it doesn't produce a
    diagnostic message in the x++ case.
    Richard Heathfield, Jun 6, 2007
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  3. x>>8 is just like x+1 or x/2 - it doesn't change the value of x.
    It gives you the value of 8, shifted right 8 places. It doesn't
    shift x itself.

    x>>=8 does what I think you are expecting.

    -- Richard
    Richard Tobin, Jun 6, 2007
  4. Chad

    Chris Dollin Guest

    Chad wrote:

    The shift operation `x >> 8` takes the value of `x`,
    shifts it 8 places to the right, and throws the
    result away.
    The post-increment operation `x++` delivers (and discards)
    the original value of `x`, and also arranges that `x`
    is incremented.
    `x++` /does/ discard its value right away, the same as `x>>8` does.
    However, the /increment operation/ is a side-effect, not a value,
    and happens regardless.
    Chris Dollin, Jun 6, 2007
  5. Why do you think that x>>8 should be like x++? Is it because of
    the repeated >? In fact >> is like +, -, *, and /, none of which
    have side effects.

    -- Richard
    Richard Tobin, Jun 6, 2007
  6. Chad

    Chad Guest

    What's the rationale behind having something like x>>8 not having a
    side-effect, but somthing like, x++ having a side effect?
    Chad, Jun 6, 2007
  7. Chad

    Chris Dollin Guest

    `x >> 8` is like `x + 1`. It has no side-effect because if it did,
    expressions would be updating their operands all over the place.

    `x++` has a side-effect because that's what it's /for/; to provide
    the value of a variable and to update it, to make common combinations
    of operations more compact (and maybe, in the old days, more efficient).

    An expression like `a[i++]` allows you to get at an element of an
    array /and/ advance the index to the next position, all in one go.
    Otherwise you'd have to find somewhere to put the increment, `i += 1`.

    Opinions on whether this kind of elegant compactness is a good idea
    are rumoured to vary. As with most programming languages features,
    it's possible to overdo things, and it's possible to misunderstand
    what such expressions actually /mean/ and where the language crouches
    ready to pull the rug out from under your feet, giggling like an
    insane ferret on nitrous oxide.
    Chris Dollin, Jun 6, 2007
  8. Chad

    mark_bluemel Guest

    What a charming image - may I reuse it, please?
    mark_bluemel, Jun 6, 2007
  9. [...]

    Um, that's not quite the way I'd put it.

    The expression x>>8 yields the value of x right-shifted by 8 bits.
    Throwing away the result isn't a feature of the ">>" operator; the
    result is thrown away because you (the OP) asked for it to be thrown
    away, by using the expression as a statement (by adding the ';').

    What x>>8 *doesn't* do is modify the value of x, just as the
    equivalent x/256 doesn't modify the value of x. Similarly, 256>>8
    doesn't modify the value of 256, and 2+3 doesn't modify the value of 2
    or 3.

    The ++ operator, as numerous others have pointed out, has the *side
    effect* of modifying the object that is its operand. That's why the
    ++ operator, unlike the >> operator, can *only* be applied to an
    object (an lvalue); 256++ is illegal.

    If you *want* to modify the value of x, replacing it with the result
    of x>>8, you can write:

    x = x>>8;

    or, equivalently:

    x >>= 8;

    A couple of notes on the original program:

    You're missing the "#include <stdio.h>". Your program may happen to
    appear to work without it, but it's mandatory if you use printf or
    anything else declared in <stdio.h>. (Somebody might point out that
    you could drop the #include and declare the printf function yourself;
    that's true, but it's a dumb thing to do.)

    You can use bitwise operators, (<<, >>, &, |, ^) on signed integers if
    you really want to, but it almost always makes much more sense to
    apply them to unsigned integers. There are rules on how these
    operators work on negative values, but I can't be bothered to look
    them up.
    Keith Thompson, Jun 7, 2007
  10. Chad

    Chad Guest

    Today, somewhere between me considering if I should make a third
    attempt to apply to UC-Berkeley and my manager at work asking me if I
    was dumb, the whole x>>8 vs x++ sank in. Then as the accounting lady
    at work was making more sexual advances at me, I realized I could have
    saved myself posting on here had I given the whole x>>8 vs x++ more
    than a millisecond of thought before giving up.
    Chad, Jun 7, 2007
  11. Chad

    Chris Dollin Guest

    Certainly. No attribution required (but appreciated if present).

    "It was the first really clever thing the King had said that day."
    /Alice in Wonderland/

    Hewlett-Packard Limited registered office: Cain Road, Bracknell,
    registered no: 690597 England Berks RG12 1HN
    Chris Dollin, Jun 7, 2007
  12. Chad

    Chris Dollin Guest

    Ooof. You're quite right, Keith; I was sloppy. Thanks for the catch.
    Chris Dollin, Jun 7, 2007
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