[DaWei]

You can not assign function addresses like this:

    void (*functionptr[3])(int *) = {void update(int *), void deleted(int *), void display(int *)};

"void update (int *)" looks like a declaration, not an address.
"update (int *)" looks like a messed up invocation, not an address.
Use the following form:

    void (*functionptr[3])(int) = {update, deleted, display};

You should not be using VC++6.0 unless there is an absolutely compelling reason to do so (your tutor has threatened to shoot you, or something). VC++6.0 was written BEFORE the C++ standard was adopted. It is non-compliant in many respects. If you use it, you will have to disregard many of the things that you have been told about producing robust and compliant code.

VC++2005 Express Edition is free for the download. If you want a CD, you only have to pay shipping. Another free compiler is Dev-Cpp, from Bloodshed. It uses MinGW, a gcc/g++ port to Windows. Neither is fully compliant with the standard (few compilers are), but they are much, much better than VC++6.0.

[Peter_APIIT]

From my knowledge, one can assign name of array name to a pointer such as

int *ptr;
int array[3];
ptr = array;            // or ptr = &array; -- Array name is constant pointer;

The same concept applied to function pointer. 
Name of function is constant pointer to that function. 
The correct methods is 
void (*functionptr[3])(int *) = {update, deleted, display};
and not
void (*functionptr[3](int *) = {update(), deleted(), dispaly()};

I just guess. I don't know whether it is correct.

By the way, i do not understand this two different code.

int *ptr;
int number = 1;
ptr = &number;

and

int number = 2;
int *ptr = number;

* and -> is deference operator - to get the value
& is to get the address.

Sorry for my stupidness.

Thanks for your help.
I will post this program afterwards in order to help someone to understand function pointer, array of structure pointer and file operation.

Your help is greatly appreciated by me and others.

[shabbir]

int *ptr; // Declaring a pointer which may point to any undefined location
int number = 1; // Declaring a variable and assigning the value 1 to it
ptr = &number; // Assigning the address of the variable number to the pointer ptr

See my comments after each line.

int number = 2; // Declaring a variable and assigning the value 1 to it
int *ptr = number; // This should be int *ptr = &number; which will have the same effect as above 3rd ststement

See my comments after each line.

Now I would suggest you create seperate thread for each of your query and having the same query in the same thread has 2 disadvantage.

1. some queries may get unnoticed.
2. Other people searching will not be helped and so may ask the same question over and over again.

[DaWei]

One can assign an array name to a pointer (myPtr = myArray) simply because the standard says that the compiler must accept that and make the conversion. It is equivalent to "myPtr = &myArray [0]". Note that this yields some apparent equivalencies in usage, however myPtr and myArray are NOT the same. When the two are dereferenced (myPtr [3] and myArray [3]), different machine code is emitted in order to produce the value in element 3. See the tutorial in my signature, the section WHAT IS NOT A POINTER, for more.

Quote:

4.2 Array-to-pointer conversion [conv.array]

1 An lvalue of type "array of N T" or "array of unknown bound of T" can
be converted to an rvalue of type "pointer to T." The result is a
pointer to the first element of the array.

2 A string literal (_lex.string_) that is not a wide string literal can
be converted to an rvalue of type "pointer to char"; a wide string
literal can be converted to an rvalue of type "pointer to wchar_t".
In either case, the result is a pointer to the first element of the
array. [Note: this conversion is deprecated. See Annex _depr_. ]
For the purpose of ranking in overload resolution (_over.ics.scs_),
....

[Peter_APIIT]

Thanks for your help.