pointers, array, functions

[Disch]

danicpp wrote:

myarray is same as &myarray[0]

<Technicality>
myarray can be implicitly cast to the same thing as &myarray[0]. However they are not the same thing -- there is a subtle difference.

myarray is an array.
&myarray[0] is a pointer to the first element.

However.. an array can be cast to a pointer to its first element. But that only happens when necessary.

This trips up newbies most commonly when they try to use sizeof:

cout << sizeof(int);   // assume this prints 4
cout << sizeof(int*);  // assume this prints 4

int array[10];
cout << sizeof(array);     // prints 40 -- the size of the array
cout << sizeof(&array[0]); // prints 4 -- the size of a pointer

int* ptr = array;     // legal assignment -- being implicitly cast
cout << sizeof(ptr);  // prints 4 -- the size of a pointer 

</Technicality>

[danicpp]

thanks cire I get it now a pointer will get a NULL value when we put a NULL value in it but still we cannot dereference it unless we don't put a valid address in it

well void I thought means nothing void written or not written have no effect(in functions at-least) but still I guess C++ recognizes nothing as void I guess, so at declaration time if we want variable to be of no type then we ought to tell c++ with void keyword

Did I get it all right???

[danicpp]

yes Disch makes sense, there is difference between array and &array[0] but array can cast this( &array[0] ) when necessary

thanks for the help, I guess I'm improving my knowledge much better way.

[Disch]

well void I thought means nothing void written or not written have no effect(in functions at-least) but still I guess C++ recognizes nothing as void I guess, so at declaration time if we want variable to be of no type then we ought to tell c++ with void keyword

You can't have a void variable -- that wouldn't make any sense.

There are 3 general uses for void that I can think of:

1) As a return type to a function:

void function()
{
}

The 'void' cannot be omitted here. This indicates that the function does not return any value.

2) To indicate that you want a function to take no parameters:

int function(void)
{
    return 0;
}

This is a C thing and not really C++.
In C++, no parameters means the function takes no parameters.
In C, no parameters means the function can take any number of parameters (similar to the C++ ellipsis)

So in C, to explicitly say the function has no parameters, you put the void in there. However this is completely unnecessary in C++.



3) void pointer:

void* ptr;

A void pointer is another C-ish concept. The idea is that it's a generic pointer that can point to anything -- the data pointed to can be of any type.

This is typically avoided in C++ because it's not type safe.

[danicpp]

The 'void' cannot be omitted here. This indicates that the function does not return any value.

but in main(){} I guess we don't need to mention void cause actually I don't and it compiles successfully.

Well but I checked right now that I cannot declare custom functions without void keyword it says ISO C++ forbids.... so
myfun(){}
void myfun(){}

void pointer!!! but when we *(ptr + 1) how does ptr know to jump 1byte or 4bytes or more bytes????

[long double main]

danicpp wrote:

but in main(){} I guess we don't need to mention void cause actually I don't and it compiles successfully.

The return type of main is int (and you must declare it as such).
So this is legal:
int main()

while these are not:
main()

void main()

long double main() // my username, lol
Some compilers will compile one or both of the first two of those above, but that doesn't mean it's legal C++.

void pointer!!! but when we *(ptr + 1) how does ptr know to jump 1byte or 4bytes or more bytes????

You can't use + with a void pointer exactly because of this.

[cire]

but in main(){} I guess we don't need to mention void cause actually I don't and it compiles successfully.

main must return type int. If your compiler compiles main(){} without complaint, it is not conformant. If you're compiling as (pre-C99) C and not C++, the return type of a function without a return type specified is assumed to be int.

void pointer!!! but when we *(ptr + 1) how does ptr know to jump 1byte or 4bytes or more bytes????

You cannot do pointer arithmetic with void pointers.

[danicpp]

thanks to all of you my mid exams are starting day after tomorrow so my cplusplus use may reduce but I will start again soon around June 17th

Thankyou all