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Comments:
<0> b = 6+6 ?
<1> Because ++a + a changes at to 6 right then in there
<1> Yes
<1> changes a *
<0> thats tricky.
<2> can sum 1 please help?
<1> Typically you wouldn't see something like that in most code, the worst you'd see is like buf[i++] = 'c' or something along those lines.
<2> ResDialog3.obj : error LNK2001: unresolved external symbol "public: virtual struct CRuntimeCl*** * __thiscall CReserve::GetRuntimeCl***(void)const " (?GetRuntimeCl***@CReserve@@UBEPAUCRuntimeCl***@@XZ)
<0> so why wont it give us a=12 already ?
<0> i mean if we do cout <<a; cout <<b;
<1> a = 12?
<1> a is just being incremented once, so it'd go from 5 to 6
<1> b would be 12
<0> ah, pardon, yes.
<0> ok now i see.
<2> in my dialog cl*** when I make an object in the OnOk handler function
<0> lets do more !
<0> this is fun.
<2> omgosh already
<2> ok I will try myself
<1> PFloyd, linker errors are usually caused by not linking in a proper .lib, or calling a function that doesn't actually have code for it (just prototyped somewhere)
<1> NrGSTaR, int a = 3; int b = a++ + ++a + a;
<0> a=4, b=12 ?
<1> a=5 b=12
<0> a++ and ++a will increase a by 2 numbers.
<1> Right.
<0> more :P
<1> I have to go actually.
<0> :P
<1> Just tell me, int c = 3; cout << c++;
<1> What's the result of that? ;)
<0> 3
<0> :)
<0> 5
<0> !
<0> 5!
<0> 4
<1> lol
<0> :)
<1> Undefined.
<1> Is what it is.
<0> why ?!
<1> That was the trick question :P
<0> because we didnt do something like b=c++ ?
<0> there has to be a statement.
<1> Oh sorry
<1> I meant cout << c++ << c; what does the second c print
<1> My bad.
<0> confused :(
<1> You were right for the one I said.
<0> :)
<0> tadaaaaaaaaa.
<1> You win.
<1> :p
<1> I have to go, good luck.
<0> so it results 5, and then undefined.
<0> right ? :) two lines.
<3> oh no he didnt...
<4> Of course I did, I'm downloading porn from your 192.168.10.10's F:
<0> so what does float rma[] = {500.0,125.0,0.5 }; mean ?
<4> An array of 3 floats
<5> same as rma[3] = {500.0,125.0,0.5 };
<0> same as float rma[1] = 500
<0> than float rma[2]=125
<0> etc ?
<5> I think float rma[2] = {125} is the valid choice
<5> rest of the array is initialized with 0
<0> I think I got so confused.
<0> :)
<5> int rma[5] = {1} means rma={1,0,0,0,0}
<0> oh, 3 numbers.
<0> ok.
<0> gotcha
<0> I think.
<0> ok, i need some help.
<0> float rma[] = {500.0,125.0,0.5 };float *ptr1 = rma; float *ptr2 = ptr1++; float num = (*ptr2) / rma[1]; cout << num << endl; num = num / *(++ptr1); cout << num << endl;
<0> I know the results, just didnt understand how they figured them out. helllllp !
<6> huh::
<6> ??
<0> what will appear in the two "cout"s
<0> :)
<0> I mean I know that 4, and 8
<0> but how come.
<0> and what does *ptr mean anyway.
<7> bleh
<8> still same problem
<8> no one wants to help me
<8> LNK2001: unresolved external symbol "public: virtual struct CRuntimeCl*** * __thiscall CReserve::GetRuntimeCl***(void)const " (?GetRuntimeCl***@CReserve@@UBEPAUCRuntimeCl***@@XZ)
<5> punkazz, use IMPLEMENT_SERIAL or IMPLEMENT_DYNAMIC or so
<5> whatever DECLARE_ you have for
<6> Punkazz is there some reason you feel compelled to be logged on 3 times
<0> hmm
<0> test
<0> great.
<0> so
<0> I know the results, just didnt understand how they figured them out. helllllp !
<0> float rma[] = {500.0,125.0,0.5 };float *ptr1 = rma; float *ptr2 = ptr1++; float num = (*ptr2) / rma[1]; cout << num << endl; num = num / *(++ptr1); cout << num << endl;
<6> NrGSTaR we have a posting website for a reason
<0> oh.
<0> that's true, I didnt realise.
<0> that's the problem with us newbies. :)
<6> it's IN the topic message
<0> Yeah, it's just that I didn't read it. Sorry. ;)
<4> Yeah, US newbies ARE problems
<9> African newbies, too
<0> You need to rewrite the topic for them.
<0> Anyway, since I already made this mistake of writing, would anyone help ? :)P
<4> What don't you understand?
<4> ptr1 = rm; // ptr1 points to the first thing in rma
<4> *ptr1 // you get that thing it points to
<0> I need it again, I didnt get it :)
<10> hey
<11> *p = &rm
<11> *p = rm // w/out & is an error, i think
<12> WOAH!
<12> It compiles!
<12> :)
<12> and works
<12> kewl
<13> Well- as long as we're talking about array implements...
<13> I made the following mistake yesterday..
<13> The {0.0, 1.1, 2.2} form may on be used when initializing an array. Otherwise you get "error: expected primary-expression before '{' token" & "error: expected `;' before '{' token"
<13> in other words- you can't do this: int twoDarray[2][2]; twoDarray[0]={5,7};
<14> Petskull and do you know why?
<13> why you can't do that?
<14> Petskull because arrays are const pointers.
<14> And arrays of arrays are const pointers to const pointers. So you can't change a const * const * using twoDarray[0]={5,7}; for example.
<13> hmmm...
<13> lemme wrap my head around that one..
<13> ... when I say 'twoDarray[0]'... I am saying '*twoDarray'?
<14> how about this: const int x = 1; x = 2;
<14> What will happen?
<13> can't do it
<13> so you're saying that in the case of arrays, it is like saying "int x=0x0060 (address in memory)"?
<14> umm no
<14> int array[2]; // same as int const * array;
<14> Except that in int array[2]; you have 2 cells of memory allocated
<14> where as in the second case its just a dangling pointer
<13> like 'int *array; array = new int[2];"
<13> ?
<14> Thats a dynamic array.
<14> Which means that it can point around to different memory locations.
<14> an array created with int array[2] will always point to the same location.
<14> Hence its a constant pointer
<5> like 'int const *array = new int[2]'
<13> ah... with an array, I can't '&array[1] = x'?
<13> hmm.... buoy....
<13> I'ma need to write that down..
<15> hin all
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