char * ( *(array_base_address + 2 ) + 3 )

tushar_pandey · 07-30-2012, 12:24 AM

char *array_base_address[] = { "roshni" , "manish" , "sona" , "bijjuda" , "ritu"};

i know that this is used for muti-dimensional array . but i have a a problem that when *(array_base_address + 2 ) is processed inside char * ( *(array_base_address + 2 ) + 3 ) , why "*(array_base_address + 2 )" is not pointing to s in "sona"

NevemTeve · 07-30-2012, 12:33 AM

Please produce a minimal but complete program, that shows the problem. (You might even try and use tags [code] and [/code] when paste it in.)

tushar_pandey · 07-30-2012, 12:47 AM

#include<stdio.h>

int main()
{
char array_base_address[][7]={"Roshni","Manish","Sona","Bajju","Ritu"};

printf("%c",*(*(array_base_address+2)+2));

return 0 ;
}

NevemTeve · 07-30-2012, 01:29 AM

Well done. And now:
1. What did you expect to happen?
2. What did happen?
3. You really couldn't use [code] and [/code] before and after the code?

Note: your program would be easier to understand with the standard operators:

Code:

printf("%c",*(*(array_base_address+2)+2));
printf("%c",*(array_base_address[2]+2));
printf("%c",array_base_address[2][2]);

the output is 'n' from 'Sona'

gchen · 07-30-2012, 05:23 AM

Quote:

Originally Posted by NevemTeve

Well done. And now:
1. What did you expect to happen?
2. What did happen?
3. You really couldn't use

Code:

and

before and after the code?

Note: your program would be easier to understand with the standard operators:

Code:

printf("%c",*(*(array_base_address+2)+2));
printf("%c",*(array_base_address[2]+2));
printf("%c",array_base_address[2][2]);

the output is 'n' from 'Sona'

support what NevemTeve said above.

Learn with each other.

: )

tushar_pandey · 07-30-2012, 05:36 AM

Sir i know i am wrong , but why not this !

printf("%c",*(*(array_base_address+2)+2)) in this line , when ever *(array_base_address+2) is processed , so after processing why this why sona term is not taking the place of *(array_base_address+2)

like

printf("%c",*(*(array_base_address+2)+2)) , so after processing *(array_base_address+2) , why it is not acting like *(sona+2) .
because *(array_base_address+2) indicates the base address of 3rd row in the array !

Sir nevemteve & gcchen

printf("%c",*(*(array_base_address+2)+2));

think about this line , we have base address array_base_address now we are adding +2 in it means now we are at 3rd row of the array , and after it , we are saying *(array_base_address+2) , it means we are dealing with sona , now we are adding +2 in sona we can say like this , *(sona+2)
.....but this is wrong , because it is multidimensional array , i know it , but why this is wrong !

NevemTeve · 07-30-2012, 06:06 AM

Please try to understand that *(array_base_address+2) is the same thing as array_base_address[2].

Note: nothing is wrong: you get the 3rd character of the 3rd row, just as you wanted.

gchen · 07-30-2012, 09:20 PM

Sorry for bad format of the contents. please see next reply below, instead of.

gchen · 07-30-2012, 09:23 PM

Code:

  1 #include <stdio.h>
  2 
  3 int main()
  4 {
  5         char *ptrnormal = NULL;
  6         char *ptroff = NULL;
  7         char (*ptranormal)[8] = NULL;
  8         char (*ptrael)[8] = NULL;
  9         char (*ptraoff)[8] = NULL;
 10 
 11         struct {
 12                 char array0[8];
 13                 char array1[8];
 14                 char array2[8];
 15 
 16         } wrap = {{'a', 'b', 'c', '\0'},
 17                   {'h', 'i', 'j', '\0'},
 18                   {'o', 'p', 'q', '\0'}};
 19 
 20         char arrayarray[3][8] = {{'A', 'B', 'C', '\0'},
 21                                  {'H', 'I', 'J', '\0'},
 22                                  {'O', 'P', 'Q', '\0'}};
 23 
 24         ptrnormal = wrap.array0;
 25         ptroff = wrap.array0 + 2;
 26         ptranormal = arrayarray;
 27         ptrael = &wrap.array0;
 28         ptraoff = &wrap.array0 + 2;
 29 
 30 
 31         fprintf(stdout, "\nptrnormal: %s\nptroff: %s"
 32                         "\n*ptranormal: %s\n*ptranormal + 2: %s"
 33                         "\n*(ptranormal + 2): %s"
 34                         "\n*ptrael: %s\n*ptraoff: %s\n",
 35                         ptrnormal, ptroff,
 36                         *ptranormal,
 37                         *ptranormal + 2,
 38                         *(ptranormal + 2),
 39                         *ptrael, *ptraoff);
 40         
 41         
 42         return 0;
 43 }

The output is:

[root@localhost test]# cc -g -Wall test.c
[root@localhost test]# ./a.out

ptrnormal: abc
ptroff: c
*ptranormal: ABC
*ptranormal + 2: C
*(ptranormal + 2): OPQ
*ptrael: abc
*ptraoff: opq
[root@localhost test]#

Hope these information is helpful for you.

: )

gchen · 08-01-2012, 03:09 AM

Please read "Expert C Programming" book (can download from internet, please google).

It contents many information about Array and pointers.

: )