Why does poo poo go clang clang?
482902
cuz you ate nails and went potty in a metal bucket?
736487
Whenever a function is called, the current context(like line no., variables value, etc.) of the execution get pushed in STACK  MEMORY area of RAM and when the function returns it pops the current context and resumes the execution. When u call the function multiple times inside another, then the stack memory may exhaust with the context value, then this is called as stack overflow.\n\nIt generally occurs when the recursion functions are poorly written.
653205
It lets the linker help the compiler piece together code that is in other files/libraries - for instance, if you do:\n\n#include <iostream>\n\nYou're going to be able to use the functions within the iostream library - otherwise, it wouldn't know where to go to get those function definitions.
11190
When you have an OO language, my suggestion will always be to use those OO features.\nThat said, for C++ I would model objects and use their constructors to get a good start in which I can determine the order of construction by calling as needed. Please see the fist link below on how to manage static construction.\nSo don't think about "functions" but "state/objects" and their "initialization".
134162
what does this have 2 do w/ hocky???
751588
Sorry.... i guess ur question a lil tricky.... Well i am C++ prog and work on the VC++ platform...
602087
First one is that the Language is Object Oriented or not .\nThe Object oriented programming languages (e.g. C++ , JAVA , .NET Languages ) are more powerful than the other programming languages and It's easy to code at the same time.\nThe second one I guess that is that language is Platform\nindependent or not . If it is platform independent then you\ncan make a choice on Operating Systems .
409704
I may not have written many programs in Java, but that sure as hell doesn't look like Java to me.\n\nEdit: It's actually C as you can see from the includes:\n#include<stdio.h>\n#include<conio.h>\n#include<math.h>\n#include<stdlib.h>\n\nC++ doesn't use <something.h> it just uses plain <something>.
356740
#: 9534 S3/Languages\n    16-Feb-91  17:33:10\nSb: #the **p syntax in 'C'\nFm: SCOTT HOWELL 70270,641\nTo: all\n\nI have seen this syntax infrequently in college 'C' text and in source code,\nbut I have yet found a good explaination.\n\n                It is in reference to pointers and I beleieve it is called\n'double indirection'??? the syntax is\n\n\n        char **p = & dummy or      char **match(p,name) What does the **\nactually say\n\nThere are 2 Replies.\n\n#: 9538 S3/Languages\n    17-Feb-91  00:35:53\nSb: #9534-the **p syntax in 'C'\nFm: Sandy Tipper 72060,76\nTo: SCOTT HOWELL 70270,641 (X)\n\nActually, it means two different but related things, depending on where\nyou find it.  In both of your examples, it was in a declaration\nstatement, (the first was complicated bu the fact that it included\nan initialization, and the second was the declaration of a function).\nthe simplest example of this type is :   char **p;  This is a declaration\nof the variable p.  The first (RIGHTmost) asterisk means that p is a \npointer.  Fair enough, but a pointer to what?  The second asterisk tells\nus that the what is a pointer itself.  So p is a pointer to a pointer.\nBut what is the pointer that p is pointing to pointing at? A char. So\np is a pointer to a pointer to char.  In your example, the initialization\nis of p, setting up p to contain the address of a variable (p points to\nthat variable).  That variable had better be a pointer to char.\nLater in the code, after the definitions and declarations have been all\ndone, the asterisk has a different meaning altogether: instead of explaining\nwhat the variable being declared is, it is an operator, "doing" something\nto the expression it is operating on.  If i is an int, and p is a pointer to\nint, and pp is a pointer to pointer to int, then follow this:\n  i = 4;\n  p = &i;\n  pp = &p;\n Question: what is **pp ?  First, read it as *(*pp)  So pp is pointing to\np, so *pp evaluates as equivalent to p.  And p is pointing to i, so *p\nevaluates as equivalent to i.  So **pp is equivalent to i.  So **pp is 4.\nSee?\nSandy\n\n#: 9539 S3/Languages\n    17-Feb-91  11:44:39\nSb: #9534-the **p syntax in 'C'\nFm: Pete Lyall 76703,4230\nTo: SCOTT HOWELL 70270,641 (X)\n\nScott -\n\n char **p = &dummy;\n\nSays: p is a pointer to a pointer that points to a char type. The &dummy says\nassign the address of dummy (i.e. a pointer) to that slot. Typically double\nindirects are used with arrays of pointer, like **argv for instance.\n\nThe 'char **match(p,name) says that match is a function that returns a pointer\nto a pointer to a type of char.\n\nPete\n\n#: 9535 S10/OS9/6809 (CoCo)\n    16-Feb-91  21:28:45\nSb: #9528-#dir file help\nFm: Everett Chimbidis 76370,1366\nTo: Pete Lyall 76703,4230 (X)\n\nNot all dirs does this happen to!! only 2 out of 32 dirs!! Can you look at this\nfor me?\n\nThere are 2 Replies.\n\n#: 9537 S10/OS9/6809 (CoCo)\n    16-Feb-91  22:05:21\nSb: #9535-#dir file help\nFm: Kevin Darling 76703,4227\nTo: Everett Chimbidis 76370,1366 (X)\n\nEv - email me your code again (and be ****sure**** you send it as text)... I\nwas going to look at it today, but the power dropped at the house and I lost it\noff my ramdisk.\n\nHave those dirs been sorted or anything, btw?  What exactly is happening? Just\ngiving the wrong dates and file sizes, or ?\n\nThere is 1 Reply.\n\n#: 9567 S10/OS9/6809 (CoCo)\n    18-Feb-91  23:18:34\nSb: #9537-#dir file help\nFm: Everett Chimbidis 76370,1366\nTo: Kevin Darling 76703,4227 (X)\n\nonly wrong on 2 out of 26 dirs!! All is wrong dates file size (name is ok)\nSending it to you!\n\nThere is 1 Reply.\n\n#: 9571 S10/OS9/6809 (CoCo)\n    20-Feb-91  08:50:45\nSb: #9567-#dir file help\nFm: Kevin Darling 76703,4227\nTo: Everett Chimbidis 76370,1366 (X)\n\nEv - your program seemed to work okay here.\n\nAre those 2 bad dirs okay?  That is, "dir e" and all the files work there?\n\nHave you checked to see ju
458776
If you're asking what is the human explanation of\neach of these declarations, I would suggest turning\nto an old Unix utility which is freely available\non the net, "cdecl".\n\nIt's a little program which converts between\nC and C++ declaration and casting syntax and\na form of natural language.  \n\n"cdecl" explains these as follows:\n\n\nint(*P(char *a))[10];\ndeclare P as function (pointer to char) returning pointer to array 10 of int\n\nint(*P(char *a[]));\ndeclare P as function (array of pointer to char) returning pointer to int\n\nint(*P[10])(char a);\ndeclare P as array 10 of pointer to function (char) returning int\n\nint*(*P)(char *a[]);\ndeclare P as pointer to function (array of pointer to char) returning pointer to int
120642
There are a few typical reasons for unresolved\nexternals in C++.  \n\nThe simplist cause is that you forgot to link in a\nrequired library.  I'm not proficient with Windows,\nbut on a Unix/Linux/Solaris machine, you can search\nthrough libraries using "nm" to find the missing\nsymbol (and then specify that library on your linker\ninvocation).\n\nMaybe the next simplist cause would be that you\nactually are missing a symbol (ie. because of an\n#ifdef or some oversight in your Makefile, some\nsource file didn't compile).  Look for the symbol\nin the source and make sure that file is correctly\ncompiled.  [By the way, it is quite possible that\nyour linker will report "mangled" names which\nincorporate the class hierarchy of the type of\nthe symbol.  You may wish to filter your linker\noutput through something to reconstitute the\noriginal symbol name.  On many systems, this filter\nis called "c++filt", but I've also seen it called\n"dem" (as in demangler).]\n\nFinally, in C++, I've seen another cause of unresolved\nsymbols being short-comings in the template instantiation\nsystem.  In the general case, the compilation system\nshould detect the use of templates and create the\nnecessary versions where and when needed, but some\ncompilers provide more limited support.  In those cases,\nit may be necessary to explicitly instantiate the \nmissing template.  Take a good look at the unresolved\nsymbol (particularly if you can "demangle" the name)\nto see if it looks like it should be provided by one\nof your (or the system's) templates.  That would be\nthe clue that this is an instantiation problem.
108454
#include "ALLOC.H"\n#include "DOS.H"\n#include "CONIO.H"\n#include "STDIO.H"\n\n\nvoid main()\n{\n char *name;\n printf(" Enter the file name...:");\n scanf("%s",name);\nplaywav(name,1.0);\n\n}\n\n\nvoid playwav(char wavefile[14],float delaytime);\nstruct WaveData {\n  unsigned int SoundLength, Frequency;\n  char *Sample;\n};\n\n\nstruct HeaderType {\n  long         RIFF;      //RIFF header\n  char         NI1 [18];  //not important\n\n  unsigned int Channels;  //channels 1 = mono; 2 = stereo\n  long         Frequency; //sample frequency\n  char         NI2 [6];   //not important\n  char         BitRes;    //bit resolution 8/16 bit\n  char         NI3 [12];  //not important\n} Header;\n\n\n\n\n\nstruct WaveData Voice;         //Pointer to wave file\n\n\nunsigned int    Base;          //Sound Blaster base address\n\n\nchar            WaveFile [25]; //File name for the wave file to be played\n\n\n/*************************************************************************\n***\n** Checks to see if a Sound Blaster exists at a given address, returns\n**\n** true if Sound Blaster found, false if not.\n**\n**************************************************************************\n**/\nint ResetDSP(unsigned int Test)\n{\n  //Reset the DSP\n  outportb (Test + 0x6, 1);\n  delay(10);\n  outportb (Test + 0x6, 0);\n  delay(10);\n  //Check if (reset was succesfull\n  if ((inportb(Test + 0xE) & 0x80 == 0x80) && (inportb(Test + 0xA) ==\n0xAA))\n  {\n    //DSP was found\n    Base = Test;\n    return (1);\n  }\n  else\n    //No DSP was found\n    return (0);\n}\n\n\n/*************************************************************************\n***\n** Send a byte to the DSP (Digital Signal Processor) on the Sound Blaster\n**\n**************************************************************************\n**/\nvoid WriteDSP(unsigned char Value)\n{\n  //Wait for the DSP to be ready to accept data\n  while ((inportb(Base + 0xC) & 0x80) == 0x80);\n  //Send byte\n\n  outportb (Base + 0xC, Value);\n}\n\n\n/*************************************************************************\n***\n** Plays a part of the memory\n**\n**************************************************************************\n**/\nvoid PlayBack (struct WaveData *Wave)\n{\n  long          LinearAddress;\n  unsigned int  Page, OffSet;\n  unsigned char TimeConstant;\n\n\n  TimeConstant = (65536 - (256000000 / Wave->Frequency)) >> 8;\n  WriteDSP(0x40);                  //DSP-command 40h - Set sample frequency\n  WriteDSP(TimeConstant);          //Write time constant\n\n\n  //Convert pointer to linear address\n  LinearAddress = FP_SEG (Wave->Sample);\n  LinearAddress = (LinearAddress << 4) + FP_OFF (Wave->Sample);\n  Page = LinearAddress >> 16;      //Calculate page\n  OffSet = LinearAddress & 0xFFFF; //Calculate offset in the page\n  /*\n      Note - this procedure only works with DMA channel 1\n  */\n  outportb (0x0A, 5);              //Mask DMA channel 1\n  outportb (0x0C, 0);              //Clear byte pointer\n  outportb (0x0B, 0x49);           //Set mode\n  /*\n      The mode consists of the following:\n      0x49 = binary 01 00 10 01\n                    |  |  |  |\n                    |  |  |  +- DMA channel 01\n                    |  |  +---- Read operation (the DSP reads from memory)\n                    |  +------- Single cycle mode\n                    +---------- Block mode\n  */\n\n\n  outportb (0x02, OffSet & 0x100); //Write the offset to the DMA controller\n  outportb (0x02, OffSet >> 8);\n\n\n  outportb (0x83, Page);           //Write the page to the DMA controller\n\n\n  outportb (0x03, Wave->SoundLength & 0x100);\n  outportb (0x03, Wave->SoundLength >> 8);\n\n\n  outportb (0x0A, 1);              //Unmask DMA channel\n\n\n\n  WriteDSP(0x14);                  // DSP-command 14h - Single cycle playback\n  WriteDSP(Wave->SoundLength & 0xFF);\n  WriteDSP(Wave->SoundLength
299395
There can be many a reasons for it\n\n1) You are trying to call up an Interrupt Request\n2) You are trying to access hardware directly.\n3) You are trying to write to a memory location that is prohibited.\n4) You may be missing some required library file.\n5) You may be throwing an unhandled system exception/error.\n\nHere you refers to your program or the native code compiler. For you have not mentioned as where exactly does the problem occur, while executing or while compiling the code.
639857
use of this pointer is to show the current referemce.\nu can pass the formal arguments and variables with same name,\n\nfor eg a part of prog\n.\n.\nvoid initialize(int x,int y)\n{\nthis.x=x;//x=5\nthis.y=y;//y=6\n}\n.\nvoid main()\n{\n..\n...\ninitialize(5,6);\n}\nif this is not used the above prog will be\n\nvoid initialize(int x,int y)\n{\nx1=x;//x1=5\ny1=y;//y1=6\n}\n.\nvoid main()\n{\n..\n...\ninitialize(5,6);\n}\n\ndid u understand..
450295
Check out this explanation\n\nhttp://homepage.mac.com/deckert/iblog/C11989312/E117193629/
471250
Fairly straight forward. The code below is the skeletal C++ code (since I am not familiar with Java).\n\nclass Complex {\npublic:\nComplex () {a = 0; b = 0;}\nComplex (float x, float y) { a = x; b = y;}\nfloat real () {return a;}\nfloat imaginary () {return b;}\nvoid Add (Complex c, Complex d) { a = c.real() + d.real();\nb = c.imaginary() + d.imaginary();}\nvoid Subtract (Complex c, Complex d) {....}\nvoid Print() { cout << a << "+ i" << b; }\nprivate:\nfloat a;\nfloat b;\n};
172063
http://www.thefreecountry.com/compilers/cpp.shtml