CODEWITHSUNDEEP
coutput
boxed__l
boxed__l

Reputation: 1336

Trouble with output in c

ptr can be assumed to have the address 1000:2000

code:

int ptr[2];
printf("%p %p %p",&ptr+1,ptr+1,ptr);

What will be the output of above code?

What I tried:
As I don't know how to interpret/convert 1000:2000 into an address, I tried manually.

I tried this by considering ptr's address as 10000+2000=12000
So, &ptr=12000 and &ptr+1=12000+sizeof(int)
ptr is the address of first element being pointed, similarly ptr+1 is the second element's address
Is this right?
How can I test this?

Upvotes: 0

Views: 88

Answers (2)

glglgl
glglgl

Reputation: 91017

This is simple pointer arithmetics if you know the types.

int ptr[2];
printf("%p %p %p",&ptr+1,ptr+1,ptr);

ptr is the array's name which decays to a pointer to the first element. Let it be 1000:2000, if it was an old book.

So ptr+1 is one element further, thus, if int has size 4, 1000:2004. (Probably int has size 2 in this book, so it would be 1000:2002.)

&ptr, however, is a pointer to the array. It points to the same place - 1000:2000 -, but to a larger item: twice as large. So adding 1, the result is incremented by the size of two integers, thus 8 resp 4. So &ptr+1 gets you 1000:2008 resp. 1000:2004 as answer.

Upvotes: 1

Lidong Guo
Lidong Guo

Reputation: 2857

+-------------------------------------+
|{ ptr[0]  ptr[1] } {ptr1[0],ptr1[1]} |
|   ^      ^           ^              |
|   |      |           |              |
|  ptr   ptr+1         |              |
|   |                  |              |
|  &ptr              &ptr+1           |
+-------------------------------------+

It is easy to understand ptr pointer the first address item of this array, ptr+1 pointer the second item of array. &ptr ,it is a pointer point a int array[2] int is a int (*x)[2] ,so &ptr +1 means the address move a int array[2].

Here is a test code:

  #include <stdio.h> 
  int main()
  {
      int ptr[2];
      int  i;
      for (i =0; i <3; ++i)
          printf("ptr + %d is :   %p\n",i,&(ptr[i])); 
      printf("\n&ptr +1 :    %p\nptr+1  :    %p\nptr  :    %p\n",&ptr+1,ptr+1,ptr);
  }

Screen Print

ptr + 0 is :    0xbf892614         <---the same as ptr        base address   
ptr + 1 is :    0xbf892618         <---the same as prt+1      4 byte add
ptr + 2 is :    0xbf89261c         < --the same as &ptr +1    4 byte add

&prt +1    :    0xbf89261c      
ptr+1      :    0xbf892618     
ptr        :    0xbf892614

Upvotes: 1

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