c programming

Question

#include<stdio.h>
int main()
{
   char *s[] = {"ice","green","cone","please"};
   char **ptr[] = {s+3,s+2,s+1,s};
   char ***p = ptr;
   
   printf("%s\n",**++p);
   
   printf("%s\n",*--*++p + 3);
   
   printf("%s\n",*p[-2]+3);
   
   printf("%s\n",p[-1][-1] + 1);
   
    
   return 0;
}

Please anyone explain this especially last printf statement.

Comments

https://gateoverflow.in/3592/gate2006-it-49

https://gateoverflow.in/120718/program-output

https://gateoverflow.in/108786/c-program

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Please let me know what will be the sizeof(ptr)?

My doubt is how many bytes of memory will be allocated to ptr? 

Some considerations:

1. char **ptr[] ; // ptr is an array of elements which are a pointer to a pointer to elements of character datatype.
2. So basically ptr is an array of pointers ( which are itself pointers)
3. And since every pointer is of unsigned int data type hence the size of ptr array should be = (no. of elements x size of a pointer) 
4. Here no. of elements is 4 and let’s assume size of a pointer ( in 64 bit system ) is 8 bytes
5. Hence size of ptr array should be 4*8 = 32;
6. If base address of ptr is 6000 then ,
   1. address of ptr[0] = 6000
   2. address of ptr[1] = 6008
   3. address of ptr[2] = 6016
   4. address of ptr[3] = 6024

Why in the answers mentioned here took addresses as 6000, 6001, 6002, 6003?

Please do provide clarity for the same.

Edit: Now I think the addresses for the elements of array s should be addressed as 5000, 5008, 5016, 5024. (If base address of s is 5000). Above arguments support the same. 

Let me know your thoughts.

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In C language the size of any object is only dependent on its type. Now, for a pointer - - it can point to any object including another pointer, size is the same - 8 bytes or 64 bits on a 64bit system. This may or may not be the same as unsigned integer on that system.

Answer 1

answer should be green of second  printf

Comments

$p[-1][-1] + 1 $

As we know a[i][j] can be written as  $*(*(a+i)+j)$

so last printf can be written as

= $*(*(p-1)-1) +1$

Now fill up the values and get the answer

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or as we know a[i]=*(a+i)

Let p[-1][-1] =B[-1] ( where B=p[-1])

B[-1]= *(B-1).................(1)

But B=p[-1]=*(p-1),so put in (1)

$*(*(p-1)-1)$

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One of my query is p is not double pointer, it will be triple pointer

So, why they are taking value of triple pointer with a double pointer?

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Are you addressing my comment or the answer? If you are addressing my comment, then I didn't get you:(

Answer 2

S

1000	2000	3000	4000
ice	green	cone	please

ptr

100	101	102	103
4000	3000	2000	1000

p

10	20	30	40
100	101	102	103

1.$**++p=*\left ( *\left (p+1 \right ) \right )=*\left ( *\left ( 20 \right ) \right )=*\left ( 101 \right )=3000$=value at 3000 is "cone"

2.$*--*++p+3=*\left ( --\left ( *\left ( ++p \right ) \right ) \right )+3$

$=*\left (--\left ( * \left ( 30\right )\right ) \right )+3$

$=*\left ( --\left(102\right ) \right )+3$ //from now 102  pointing to 3000

$=*\left ( 101 \right )+3=3000+3=3003=$will print "e"

3.$*p\left [ -2 \right ]+3=*\left ( *\left ( p-2 \right ) \right )+3=*\left ( *\left ( 10 \right ) \right )+3=*\left ( 100 ) \right )+3=4003$=will print value at address 1000="ase"

Now pic is

S

1000	2000	3000	4000
ice	green	cone	please

ptr

100	101	102	103
4000	3000	3000	1000

                                                              p

10	20	30	40
100	101	102	103

means here p pointing to value at address 30 and address 102 pointing to 3000 too

4.$p\left [ -1 \right ]\left [ -1 \right ]+1$

Now p is pointing to "cone"

$p\left [ -1 \right ]\left [ -1 \right ]$ will print 1 row back of it.i.e. "green"

$=*\left ( 2000 \right )+1=2000+1=2001$=it will print   "reen"

Comments

I think you are making a mistakes in second printf statement.

char ***p = ptr; $\rightarrow$ It is pointer to a pointer to a pointer to a character. 

Here you have mention it as

p

10	20	30	40
100	101	102	103

It seems like wrong.

It should be like this

p$\rightarrow$ address should be 10 and value should be 100 only. 

It wont print anything in output. I think Probably your all answer is wrong.

As you are making mistakes in block diagram of char ***p = ptr;

Check the block diagram of @abhishekmehta4u , that is the correct block diagram of this problem.

Better you should compile the program and check the output.

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mam ,i thought your second printf statement is not correct

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I think in the 2^nd Output your analysis is wrong because it will print no output according to me.

Answer 3

I think this should be the answers printed by printf

Comments

ptr[0] is pointing to s+3

but why u took reversing the address?

means u took ptr pointing to s

and p is pointing to ptr+3

Am I rt?

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ptr[0] has base address of  s+3 and intially p is pointing to base address of ptr