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  1. If I have a 2d array B defined as :

    int B[2][3] = {{1,3,5},{2,4,6}};

    Is int **p = B same as int (*p)[3] = B ?

  2. int **f = B; printf("%d ",*f+1);

    gives 5 as output while printf("%d ",*f) gives 1 as answer. Why is that happening?

  3. printf("%d ",**f);

    returns a segmentation fault! Why?

Dubby
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2 Answers2

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  1. No. int **p = B; is an error. (Both a compilation error, and a logical error). An int ** must point to an int *. However, there are no int * stored in B. B is a group of contiguous ints with no pointers involved.

  2. int **f = B; must give a compilation error. The behaviour of any executable generated as a result is completely undefined.

  3. See 2.

To explain why you might be seeing 1 and 5. (The C standard does not define this, but your compiler bulls on ahead anyway). Probably your compiler treats the line as

int **f = (int **)B;

Then the expression *f will read bytes from the storage of B (which actually hold ints) and pretend that those are the bytes that make up a pointer representation. This is further undefined behaviour (violation of strict-aliasing rules). Probably the result of this is that *f is a pointer to address 0x00000001.

Then you print a pointer by using %d, causing further undefined behaviour. You see 1 because your system uses the same method for passing int to printf as it does to pass int *.

When you add 1 to (int *)0x00000001, you get (int *)0x00000005, because incrementing a pointer means to point to the next element of that type.

When you dereference this pointer, it causes a segfault because that address is outside of your valid address space.

M.M
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  • on C99 it just gives a warning `warning: initialization from incompatible pointer type [enabled by default]` – Dubby Aug 01 '14 at 12:40
  • @Dubby the technical term is *a diagnostic*, and the program is ill-formed. It's up to the individual compiler if it wants to proceed after that (and the behaviour has left the realm of what is covered by the C standard) – M.M Aug 01 '14 at 12:42
  • So basically a declaration like `char *argv[]` is not a 2 dimensional array i.e. not the same as `char argv[m][n]` ? – Dubby Aug 01 '14 at 12:43
  • @Dubby; Yes. Its not the same. Pointers are not arrays. – haccks Aug 01 '14 at 12:44
  • No, that's a function parameter and the meaning of those is different to variable declarations. [see here](http://stackoverflow.com/questions/22677415/why-do-c-and-c-compilers-allow-array-lengths-in-function-signatures-when-they/) for details. – M.M Aug 01 '14 at 12:45
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1) Is int **p = b same as int (*p)[3] = b ? - No. int **p = b is an error.

Because here int **p is a pointer to pointer to an integer, but int (*p)[3] is pointer to an array of 3 integers!

2) int **f = B; It is an error, May results in Undefined behavior!

3) printf("%d ",**f); - It is same as (2). int **f = B; is error, so Undefined behavior!

NOTE: To avoid this type of error enable some warning flags in compiler option and try!

Sathish
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