C pointer to array array of pointers disambiguation

Question

What is the difference between the following declarations   int  arr1 8   int   arr2  8   int   arr3 8      What is the general rule for understanding more complex declarations

User · Answer

The answer for the last two can also be deducted from the golden rule in C:

Declaration follows use.

int (*arr2)[8];

What happens if you dereference arr2? You get an array of 8 integers.

int *(arr3[8]);

What happens if you take an element from arr3? You get a pointer to an integer.

This also helps when dealing with pointers to functions. To take sigjuice's example:

float *(*x)(void )

What happens when you dereference x? You get a function that you can call with no arguments. What happens when you call it? It will return a pointer to a float.

Operator precedence is always tricky, though. However, using parentheses can actually also be confusing because declaration follows use. At least, to me, intuitively arr2 looks like an array of 8 pointers to ints, but it is actually the other way around. Just takes some getting used to. Reason enough to always add a comment to these declarations, if you ask me :)

edit: example

By the way, I just stumbled across the following situation: a function that has a static matrix and that uses pointer arithmetic to see if the row pointer is out of bounds. Example:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define NUM_ELEM(ar) (sizeof(ar) / sizeof((ar)[0]))

int *
put_off(const int newrow[2])
{
    static int mymatrix[3][2];
    static int (*rowp)[2] = mymatrix;
    int (* const border)[] = mymatrix + NUM_ELEM(mymatrix);

    memcpy(rowp, newrow, sizeof(*rowp));
    rowp += 1;
    if (rowp == border) {
        rowp = mymatrix;
    }

    return *rowp;
}

int
main(int argc, char *argv[])
{
    int i = 0;
    int row[2] = {0, 1};
    int *rout;

    for (i = 0; i &lt; 6; i++) {
        row[0] = i;
        row[1] += i;
        rout = put_off(row);
        printf("%d (%p): [%d, %d]\n", i, (void *) rout, rout[0], rout[1]);
    }

    return 0;
}

Output:

0 (0x804a02c): [0, 0]
1 (0x804a034): [0, 0]
2 (0x804a024): [0, 1]
3 (0x804a02c): [1, 2]
4 (0x804a034): [2, 4]
5 (0x804a024): [3, 7]

Note that the value of border never changes, so the compiler can optimize that away. This is different from what you might initially want to use: const int (*border)[3]: that declares border as a pointer to an array of 3 integers that will not change value as long as the variable exists. However, that pointer may be pointed to any other such array at any time. We want that kind of behaviour for the argument, instead (because this function does not change any of those integers). Declaration follows use.

(p.s.: feel free to improve this sample!)

User · Answer

I guess the second declaration is confusing to many  Here s an easy way to understand it   Lets have an array of integers  i e  int B 8    Let s also have a variable A which points to B  Now  value at A is B  i e    A     B  Hence A points to an array of integers  In your question  arr is similar to A   Similarly  in int    C   8   C is a pointer to an array of pointers to integer

User · Answer

typedef int   PointerToIntArray     typedef int  ArrayOfIntPointers

User · Answer

Here s an interesting website that explains how to read complex types in C  http   www unixwiz net techtips reading-cdecl html

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In pointer to an integer if pointer is incremented then it goes next integer   in array of pointer if pointer is incremented it jumps to next array

User · Answer

I think we can use the simple rule     example int     ptr       start from ptr      ptr is a pointer to    go towards right   its     now go left its a     come out go right       so   to a function which takes no arguments   go left  and returns a pointer   go right  to  an array  go left   of integers

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int  arr 8      An array of int pointers  int   arr  8      A pointer to an array of integers   The third one is same as the first   The general rule is operator precedence  It can get even much more complex as function pointers come into the picture

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int  arr1 5    In this declaration  arr1 is an array of 5 pointers to integers  Reason  Square brackets have higher precedence over    dereferncing operator   And in this type  number of rows are fixed  5 here   but number of columns is variable   int   arr2  5    In this declaration  arr2 is a pointer to an integer array of 5 elements  Reason  Here     brackets have higher precedence than      And in this type  number of rows is variable  but the number of columns is fixed  5 here

User · Answer

I don t know if it has an official name  but I call it the Right-Left Thingy TM    Start at the variable  then go right  and left  and right   and so on   int  arr1 8     arr1 is an array of 8 pointers to integers   int   arr2  8     arr2 is a pointer  the parenthesis block the right-left  to an array of 8 integers   int   arr3 8      arr3 is an array of 8 pointers to integers   This should help you out with complex declarations

User · Answer

Here s how I interpret it   int  something n        Note on precedence  array subscript operator      has higher priority than   dereference operator        So  here we will apply the    before    making the statement equivalent to   int   something i         Note on how a declaration makes sense  int num means num is an int  int  ptr or int   ptr  means   value at ptr  is   an int  which makes ptr a pointer to int    This can be read as   value of the  value at ith index of the something   is an integer  So   value at the ith index of something  is an  integer pointer   which makes the something an array of integer pointers   In the second one   int   something  n     To make sense out of this statement  you must be familiar with this fact      Note on pointer representation of array  somethingElse i  is equivalent to   somethingElse   i    So  replacing somethingElse with   something   we get    something   i   which is an integer as per declaration  So    something  given us an array  which makes something equivalent to  pointer to an array

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Use the cdecl program  as suggested by K amp R       cdecl Type  help  or     for help cdecl gt  explain int  arr1 8   declare arr1 as array 8 of pointer to int cdecl gt  explain int   arr2  8  declare arr2 as pointer to array 8 of int cdecl gt  explain int   arr3 8   declare arr3 as array 8 of pointer to int cdecl gt    It works the other way too   cdecl gt  declare x as pointer to function void  returning pointer to float float    x  void

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int  a 4      Array of 4 pointers to int  int   a  4     a is a pointer to an integer array of size 4  int   a 8   5     a is an array of pointers to integer array of size 5

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As a rule of thumb  right unary operators  like         etc  take preference over left ones   So  int    ptr       would be a pointer that points to a function that returns an array of pointers to int  get the right operators as soon as you can as you get out of the parenthesis

[c] C pointer to array/array of pointers disambiguation

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