hi,

since I know java and pascal I have som problem with pointers.

In C++ It seams that I get a pointer and not memory referance when using array. With a cheracter I do not need a * (pointer ref) but for the array I need it since I get a invalid conversion from `char*' to `char'.


This prog compiles (gcc)

void function1(char chr);

main ...
{
char chr = "";
function1(chr);
}

void function1(char chr)
{}

This one as well

void function1(char * arr);

main ...
{
char array1[10] = "123456789";
function1(array1);
}

void function1(char * arr)
{}

This one givs error invalid conversion from `char*' to `char `

void function1(char arr);

main ...
{
char array1[10] = "123456789";
function1(array1);
}

void function1(char arr)
{}


SO I think the syntax isn't clear. Does someone have an explication why its like this and not made with an pointer ref like: char * array1[10] = "123456789"; How do I get the memory alocation without a pointer? If its possible.


I love the vector class i java is it some thing like it in C++?

I will try and explain this in terms of what C does for the following declarations:

char ch; This sets a side the amount of space for 1 character, exactly what you would expect.

char array[10]; This creates a pointer to a char (char *) called array. It points to a statically allocated block of 10 characters. Array is a pointer to the address where this block of memory starts.

So say this is your memory space being mapped by array[10] = "123456789":

Address Value
0x0010 1
0x0011 2
0x0012 3
....
0x0019 4

All array is is a pointer to 0x0010. To get the value at that address you would type *array, or array[0]. Typing array[x] is equivalent to typing *(array + x).

You can't allocate memory without a pointer in C. Pointers are very powerful, but take some getting used to. If you want to look into dynamically allocating memory you would do something like this:

char *array = malloc(sizeof(char) * 10); //points array to a block of memory that can hold 10 char.

array = realloc(sizeof(char)* 20); //reallocates array to point to a block of 20 characters

free(array); //releases the memory pointed to by array.

You also probably want to check out the memcpy command.

One thing you have to be really careful is your bounds, C will let you go beyond array bounds without complaining at compile time. However, those kind of situations often cause unusual things happening in your program or the dreaded words "segmentation fault".

Thanks for the expanation. I got it now and its just to accept the notation. Even if I think its clerar if it would be
char * array[10] = "123456789";

Ah, but what you have written there is a pointer to a pointer that points to a static block of memory big enough for 10 characters

blah[] and *blah are equivlent
so *blah[] and **blah and blah[][] are equivlent.

Well I been thinking about this and then I found some info in kdevelop under GCC compiler error messages. To add the [] in method head gives a nice syntax.


warning: passing arg 1 of `cpystr' makes integer from pointer
without a cast

This is the code causing the problem:
void cpystr( char item);
main()
{
char src[]="martin leslie";
cpystr(src);
}
cpystr(char item)
{
}

It should be....
void cpystr( char item[]);
main()
{
char src[]="martin leslie";
cpystr(src);
}
cpystr(char item[])
{
}

Actually jtshaw got it a little wrong.

When you have:

char arr[10];

It actually allocates 10 bytes. No pointer is allocated.

When an array is passed as a paramter to a function, the
address of the first character is sent to the function. This
is done so the C language doesn't have to track sizes of
arrays and copy all the bytes onto the stack for a function
call.

Also, people rarely use the [] to denote arrays being passed to
a function. You'll see (and people will generally expect) *
in most cases.

Good luck!

-Craig