I understand that it is a command to exit the program, but what I *don't* understand is why I see a number inside the () as a parameter. In some examples, I see people calling the function and it looks like this: "exit(8);". What is the 8 doing inside the parentheses...?
129048
It is a return code from the program.  The shell that ran the program can check to see if the program exited with an error or normally.  I believe return 0 is for normal exit, and return anyother number is for an error.
39998
A "protoype" is a statement that simply tells the program the format of a function call.\n\nLet's say that you create a function that adds up two doubles:\n\ndouble Add (double A, double B)\n{\n   return A + B;\n}\n\nNow in the header file associated with this function you would insert a function prototype:\n\ndouble Add (double A, double B);\n\nwhich you will notice looks very much like the function declaration itself.\n\nWhy do you need to do this? It is simple actually. If you were to call the function "Add" in another file, this other file does not know what types the parameters are, and what type the return value is. In order for the compiler to do its job, it needs to know this information.\n\nSo you simply "#include" the header file with the prototype, and the prototype does not actually perform anything other than telling the compiler that the function has two parameters, both of type double, and that the function returns a double.\n\nHope this helps!
580423
When variables are passed to a procedure or function they can be IN, OUT or INOUT.  An IN variable is passed into the procedure or function and is used, but can not be changed.  An OUT variable is passed to the procedure but it can be changed and left in the changed state when the procedure ends. \n\nAn INOUT variable is passed to a procedure or function, and it can be used by the block, changed by the block, and left in a “changed” state when the block ends.  A function can only be passed an IN variable and must return a variable.  If this is confusing, don’t worry.  When we get to name blocks and provide some examples it will be much cleare
481398
what do you mean by parent function, can you clarify???????
224216
Delete (when used with a pointer) destroys what the pointer points to (e.g., pointer is still there, and can point to something else).  Destroy destroys the pointer/object itself.\n\nFor instance:\n\nchar** charArrayPtr = new char[50];\ndelete charArrayPtr;\ncharArrayPtr = NULL;\ncharArrayPtr = new char[25];\n\nBut if you did\n\ndestroy(charArrayPtr);\n\nyou couldn't use charArrayPtr anymore.\n\nReally, no real need to call destroy explicitly, for the most part.
569472
ok its like this...() or parenthesis meant multiplication. Then multiplyin negative by negative is = to +. So now its gonna be +a or simply a
125484
You need to think about your exit return.\n\nfunction yourfunction(isfinished)\n{\nif (isfinished==false)\n{\ndo something if it's done isfinished=true\nyourfunction(isfinished);\n}\n}\npublic void main()\n{\nbool isfinished='false';\nyourfunction(isfinished);\n}\n\nSomething like that.
73445
Lets examin the first output case you have, f.fact(3):\nThe first layer of this function call will have result = fact(3-1)*n.  This will of course call another layer of the fact function with result = fact(2-1)*n, and this will then call fact on 1. \n\nSince the base case for the recusion is 1, 1 will be returned.  And the function call will return values in reverse order.  So result = fact(2-1)*n will no be result = 1*2.  Which will then be returned to the function that called it, result =fact(3-1)*n is now result = 2*3. and that value is finally returned.\n\nThe key to understanding recursion is knowing what the basecase is.  If you can find the base case then evaluating the recursion is simply.  Just follow the function all the way through until you hit that base case, then work your back out example... 1 -> 2 -> 3 -> 4  then to get out we'd go 3 -> 2 -> 1.\n\nHopefully this helps.
503288
javap -c \nprints out "diassembled" java code. Each line, that starts with a number, followed by a colon is an "instruction". Just count those lines.
81191
void anyfun(void);\nvoid anyfun();\n\nTechnically, the above are the same. The compiler will assume an empty parameter list to mean (void). Thus, the above are the same. Style-wise, the latter of the two is preferred for C++ programming.\n\nvoid anyfun(void);\nanyfun();\n\nThe compiler will assume you want to have no parameters, but it will also *assume* you will return a value of type "int". The compiler will have to do something for the return since you did not specify a type, and it chooses int as the return type for the function. This is just not good style and should be avoided in all cases.\n\nvoid anyfun(void);\nanyfun(void);\n\nAgain, the compiler is going to respect your (void) parameter list, but will supply the default "int" return type. This should be avoided in all cases. It is always good style and proper programming practice to *always* specify the return type of a function, especially if you don't intend to reutrn anything.\n\nYour first function declaration:\n\nvoid anyfun(void);\n\nis the best style and most commonly accepted style-wise. Always specify the return type of a function. Parameters are optional and () is proper style in C++ if you don't have parameters.
300035
I have to confess, I can't seem to make sense of your question.  Looking at question (i), how can one formula give both 7420 and 3392 as an answer?\n\nAre you talking about functions?  As in, we have some formula, and when we give it 7420 as an "input," the resulting value is 3392?  If so, you'd write that f(7420)=3392.  But I'm not sure that's what you mean.  Can you give us more detail?
37744
I rember doing something like this: you had to take the int you store the input in, and divide it by ten with the mod operator (%), and store it in a new int.  Since the mod operator divides the two numbers, and then returns the remainder, that will separate your last number.  For example:\n\nSay the input is 4321, and you have it stored in int input;\n\nint first_num = input % 10;\n\nthis takes your value in input, divides it by 10, then takes the remainder and stores it in first_num.  In this case, first_num should now equal 1.  \n\nAfter that, you have to shorten your input from 4321 to 432 by dividing (regular dividing this time) it by ten, and storing it in another new int.  (Since int can't have any decimals, dividing by ten just shortens it by the first number).  Then you do the whole thing over until you have all four numbers in separate ints, and then output them to separate lines.  \n\nHope that's clear enough--it's hard to explain this stuff so it's easy to understand.  Good Luck!
350914
It means that you distribute (multiply) what is outside of the parentheses to each term within the parens..\n\nAre you finished with the little game you're playing, by the way?
367674
The signature of a function include it's return type, it's arguments types, it's modifiers, and it's name.  \n\nIt must be unique in a given scope and is the way it is identified by the compiler.  The way it write itself as a unique entity, so by analogy it's called a signature.
572058
Excuse me for my poor C knowledge but ATOL already converts a string of numbers into an actual number. Therefore if you take "strlen" of ATOL, you're trying to calculate the length (# of digits) of a number instead of a string, which strlen will only accept.
297938
Above answer is ok except for not using the -- operator:\nNeed to start at 10 and to decrement i use the statement:\n\ni--;\n\nThis changes the problem somewhat.  Another problem in the first answer is that fact is not initialized, so it will always equal zero resulting in a division by zero error.
141072
It usually should have called the constructor A() if you are making the object in maner "A a1;" in the main function. But also you have overloaded the operator "()". SO, it makes a confusion and actually none of the output from the above constructors are called. Another thing, its not a good thing to inital values in parameters. I've often faced these problems when I overloaded the "()" operator. For any further questions you can contact me.