A question about Arrays and Pointers

Hey,

I am wondering what this is:

Code:

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int* a[3]

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- Is it a array of int* or
- is it a pointer to an array of ints?

I am asking because you can actually deepcopy multidimensional arrays using memcopy, while you cannot using type**.

After some research I experienced that an multidimensional array is stored as a coherent "object". There are no pointers to adresses stored.

So I am wondering what exactly happens when I pass an array as an type**.
Does C create pointers to actualy locations, because if not: How does the computer know the size of the stored arrays?

Thank you!

"a" is a static array of "int*" and of size "3".

A multi-dimensional array in C is just 1-D in memory. For example, the following two arrays are equivalent:

Code:

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#define COLUMNS 3
#define ROWS 4

int a[ROWS][COLUMNS];
int b[ROWS*COLUMNS];

a[2][1] = 123;
b[2*COLUMNS + 1] = 123;

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Those are the so-called 2-D arrays. What you have here is an array of pointers. Sure, it can be used as a 2-D array, but don't assume that they are equivalent.

Quote:

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because if not: How does the computer know the size of the stored arrays?

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It doesn't.

This is a 3 element array of pointer to ints

Quote:

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Code:

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int* a[3];

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Used like this...

Code:

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int *a[3]; /* Notice that the * is usually put next to the variable name */
int x,y,z;
x = 1;
y = 2;
z = 3;
a[0] = &x;
a[1] = &y;
a[2] = &z;

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If you want to create a pointer to an array of int

Code:

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int a[3] = {0};
int (*b)[3];
b = &a;

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For more info on declarations...
Reading C Declarations
http://users.ece.utexas.edu/~ryerrab...eclPrimer.html

Quote:

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How does the computer know the size of the stored arrays?

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If it is defined in scope you can use

Code:

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int apple[3] = {0};
size_t appleSize;
appleSize = sizeof(apple)/sizeof(apple[0]);

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Otherwise you need to pass the size of the array into the functions that you wish to use, like the second argument in fgets.

As everybody already answered, this:

Code:

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int *a[3];

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Is an array of 3 pointers to int. And this:

Code:

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int (*a)[3];

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Is a pointer to an array of 3 ints.

This:

Code:

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int **a;

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Is a pointer to a pointer which points to an int.

The three declarations are completely different from each other.

In the context of a function, main(), for exemple, the argument argv can be writen as one of these two:

Code:

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int main( int argc, char *argv[] )
int main( int argc, char **argv )

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They are equivalent because any array in the argument declaration is passed as a pointer. So, if you have something like this:

Code:

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// array of 3 ints.
int v[3] = { 0, 1, 2 };

// array of 3 pointers to ints.
int *a[3] = { &v[0], &v[1], &v[2] };

// Prototype. accepts a pointer to a pointer to int.
int f( int **pp );

// Calls f() with a pointer which points to pointers to ints.
f( a );

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Here a is converted to a pointer to the first element of the array (identified by the symbol a). Which is the same as:

Code:

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f(&a[0]);

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In the context of memset(), for example, the pointer in the argument points to a buffer which will be filled. A pointer of type int ** will be a pointer which points to an array of pointers, so the pointers pointed will be filled. Let's say:

Code:

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memset( ( void * )a, 0, sizeof a );

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This will fill the array a with zeroes, so a[0] = a[1] = a[2] = 0 (which is the same as NULL). Notice the array v remains untouched.

Understood! Thank you!