If I have a 2d array B defined as :
int B[2][3] = {{1,3,5},{2,4,6}};Is
int **p = Bsame asint (*p)[3] = B?int **f = B; printf("%d ",*f+1);gives
5as output whileprintf("%d ",*f)gives 1 as answer. Why is that happening?printf("%d ",**f);returns a segmentation fault! Why?
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yeah! its a typo. Corrected itDubby– Dubby2014-08-01 12:42:11 +00:00Commented Aug 1, 2014 at 12:42
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That's a pointer to a 1D array, not a pointer to a 2D array.Qaz– Qaz2014-08-01 12:51:11 +00:00Commented Aug 1, 2014 at 12:51
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This may be a bit helpful.ST3– ST32014-08-01 13:07:44 +00:00Commented Aug 1, 2014 at 13:07
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2 Answers
No.
int **p = B;is an error. (Both a compilation error, and a logical error). Anint **must point to anint *. However, there are noint *stored inB.Bis a group of contiguousints with no pointers involved.int **f = B;must give a compilation error. The behaviour of any executable generated as a result is completely undefined.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.
5 Comments
Dubby
on C99 it just gives a warning
warning: initialization from incompatible pointer type [enabled by default]
M.M
@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)
Dubby
So basically a declaration like
char *argv[] is not a 2 dimensional array i.e. not the same as char argv[m][n] ?
haccks
@Dubby; Yes. Its not the same. Pointers are not arrays.
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!
