Why these pointers are showing different addresses?

Question

For the following code fragment:

 unsigned int *ptr[10];
 int a[10]={0,1,2,3,4,5,6,7,8,9};
 *ptr=a;
 printf("%u %u",ptr,a);

i checked on codepad.org and ideone.com.On both compilers its showing different values of ptr and a

These 2 are different that's why showing different addresses — Omkant, Dec 17 '12 at 09:01
a similar question just answered [here](http://stackoverflow.com/questions/13910749/difference-between-ptr10-and-ptr10) — Grijesh Chauhan, Dec 17 '12 at 09:01

score 1 · Answer 1 · answered Dec 17 '12 at 09:02

1

With warnings on:

pointer targets in assignment differ in signedness
format ‘%u’ expects type ‘unsigned int’, but argument 2 has type ‘unsigned int **’
format ‘%u’ expects type ‘unsigned int’, but argument 3 has type ‘int *’

answered Dec 17 '12 at 09:02

David Ranieri

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score 0 · Answer 2 · answered Dec 17 '12 at 09:03

When used in pointer context, ptr points to the beginning of ptr array, while a points to the beginning of a array. These are two different arrays that occupy completely different places in memory. Why would they be the same?

Of course, printing pointer values with %u is a crime. Use %p. That's what %pis for.

score 0 · Answer 3 · answered Dec 17 '12 at 09:04

0

This is a array of pointers

*ptr[10]

If you want to assign a to this go:

(*ptr)[10]

answered Dec 17 '12 at 09:04

Theolodis

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score 0 · Answer 4 · answered Dec 17 '12 at 09:05

uint *ptr[10] is equivalent to uint **ptr and assignment *ptr = a is the same as ptr[0] = a which assign a to first offset inside ptr array, it doesn't touch value of ptr itself...

You maybe wanted to used one of these:

ptr = &a;

// Or
printf("%u %u",ptr[0],a);

// Or
unsigned int *ptr;
ptr = a;

Why these pointers are showing different addresses?

4 Answers4