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Comments:
<0> nope it should be something else
<1> melfar did tell you what it was.
<0> yeah, but what's bitwise ? :)
<1> It means it operates on the bits.
<2> and it's wise
<0> wait, if I do a>>1, what would the result be ?
<3> NrGSTaR, it helps if you think of variable values in terms of their binary representations.
<1> 6 = 0x06 = 00000110.
<1> 6 >> 1 = 0x06 >> 1 = 00000011
<0> oh.
<1> It shifts the bits to the right.
<0> so a>>1 will be 0x0a basically
<1> No.
<1> a>>1 will be whatever value is in a, shifted one bit to the right.
<0> binary.
<0> and the result (cout << a;) will be binary or back to normal ?
<3> It'd be the regular representation
<1> Errrr...
<0> I guessed so
<0> got it :)
<3> Using bitwise operators doesn't change the way it's represented using print functions and such
<0> very simple :) althoug why would i want to use it ?
<1> I think you'Re confusing things.
<3> It comes in handy for things such as hardware interactions, cryptography, etc.
<1> cout is a stream. It overrides the bitwise operators. It does not do bitshifting.
<0> oh.
<0> yeah.
<0> yes, i meant in a function
<3> Oh good eye DrkMatter
<1> What I was telling you about are the effects of << >> with integer values.
<1> int a=4; cout << a; a = a >> 1; cout << a;
<0> like.. a=6, b=a<<1; cout <<b;
<1> This will output: 4 2.
<0> yep
<0> got it :)
<1> What you jsut said would output 12.
<3> You can do cout << (a << 1); I believe.
<0> also
<0> :)
<0> can i ask more ?
<1> You sure can.
<1> We answer you long time.
<3> lol
<0> &
<0> :)
<3> Might I suggest you read: http://www.cprogramming.com/tutorial/bitwise_operators.html
<0> & is bitwise related too ?
<3> By itself, yes.
<0> ^
<3> & >> << | ^
<0> and |
<0> also ?
<3> Yes.
<0> ah
<0> great. now that makes things simple :)
<3> Just remember that && is different from & and such.
<0> oh ?
<2> & is the worst thig in c/c++
<2> thing.
<2> it has 3 distinct and unrelated meanings
<3> && is logical AND while & is bitwise.
<1> And ~. Don't forget ~.
<1> & has 3 distinct meanings&
<1> ?
<0> these are all related with what they call as "operators" right ?
<1> What would they be, except from bitwise and?
<3> Yes NrGSTaR
<0> :)
<0> Pretty simple, only I guess I'll have to look for excersises from the web.
<0> that makes the best practice. instead of theory.
<0> you guys, hmm think that I could learn functions in a few hours ? :)
<0> its not that complicated if explained well, no ?
<2> DrkMatter: let's see, taking an address maybe ?
<2> and for references
<2> NrGSTaR : all of these special symbols that have meanings (including +, ==, etc) are called operators
<2> they are actually simple just function, with a nice way to call them
<0> ok
<0> so
<0> a++ would be a+1, right ?
<2> not quite
<2> a++ both increases a, and returns it's value before the increase
<2> so b = a++ is the same as b = a; a = a+1;
<2> just a++ is the same as a+=1 which is the same as a=a+1
<0> a=1; a++, would result 2?
<2> after both expressions the value of a would be 2, right
<2> here's a tricky one : a = 1; a+= a++ + ++a;
<0> lets see
<2> it's not quite defined in the standard what the value of a should be after this executes
<2> so the point is - don't write stuff like that if you want to be sure what the program does :)
<0> 4 ?
<2> no way it can be 4, at least 5
<0> a++ would make 2.
<0> then ++a would make what ?
<2> but it doesn't matter - prefer simplier and more readable code
<2> why do you ***ume it's calculated from left to right ?
<0> I would, but when you're tested, they always make it treaky
<0> ++a would be 1+a ?
<2> ++a increases a, and returns the increased value.
<2> a++ increases, but returns the previous value
<2> a = 1; b = a++; // b = 1; a = 2
<2> a = 1; b = ++a; // b = 2; a = 2
<0> confusing. I'll read it a couple of more times :P
<0> and then you can ask me a treaky question and i'll see if i got it
<0> lets try
<0> :)
<0> ask me a treaky one :P
<3> A really tricky one? :P
<3> Or just tricky?
<0> lets start from tricky
<0> and go on to the real one :P
<3> The real one is a trick question actually, :P
<3> int a = 5; int b = a + a++;
<3> What's b?
<0> b=11 ?
<3> 10 :p
<0> or 10
<0> yes, because the value is the same, we didnt actually do anything.
<3> a will then be 6 after that statement though.
<3> But only after b= a + a++;
<3> What about b = a++ + a; ?
<0> b=6+5=11
<0> ?
<4> allo
<3> It's 10 still
<3> :P
<3> Here's a way of thinking about it, if you see a++, it won't actually change a until the statement is finished (look for a ;)
<0> oh, so it was backwards.
<0> but once the statement is finished a becomes a+1
<3> Yes
<0> a=2, b=a++;
<3> b = 2, a = 3 after
<0> will result a=2, b=3, or b=a (which is 2) and a=3
<0> ah
<0> :)
<5> can someone help?
<0> and a=2 b=++a, will result both to be 3
<3> Yes.
<3> Now how about int a = 5; int b = ++a + a;
<3> This one is tricky.
<0> a=11, b=11
<3> Actually, a = 11, b = 12
<0> ah
<3> Or sorry
<0> yes, ok i think i got it.
<3> a = 6
<3> a = 6, b = 12
<0> yes :)
<0> b = 12
<0> ?
<3> Yes
<0> b = 6+6 ?
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