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Character array..........

Posted on 2003-10-30
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Last Modified: 2010-04-15
Hi,
 
  When i compiled the following program, I got "9" as output. I couldnt able to understand the line "p=(buf+1)[5]".  What this line does?

 main()
{
  char *p;
  char buf[10] ={ 1,2,3,4,5,6,9,8};
  p = (buf+1)[5];
  printf("%d" , p);
}

Thanx.
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Question by:hemanexp
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by:Sys_Prog
Comment Utility
buf gives you the base address of array. i.e. address of first elemnet, in your case address of 1.

Now (buf + 1) would give you address of 2nd location i.e. 2

And then (buf+1)[5] means 6th location from (buf+1)

And u are assigning the value at (buf+1)[5] to p

Thus output
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dennis_george earned 20 total points
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hi,

 buf  --> address(pointer to) of the first value of the array

 so (buf  + 1 ) --> address of the second member of array (since array data is stored in continuous memory)

 an array (buf[x])  can also be represented by

  buf[x] --> *(buf + x) --> *(x + buf) --> x[buf] ;

 so

 (buf + 1)[5] -- > *( (buf +1) + 5) --> *(buf + 6) --> buf[6] --> 6[buf]

 Hope you got this

Dennis
 
 
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by:dennis_george
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by:Ajar
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char *p;
char buf[10] ={ 1,2,3,4,5,6,9,8};

//buf[0]    buf[1]      buf[2]      buf[3]     buf[4]   buf[5]  buf[6]    buf[7]  .....
//SAME AS
//buf+0    buf+1      buf+2        buf+3     buf+4    buf+5  buf+6    buf+7  .....
//1          2             3              4            5           6         9         8


//so       (buf+1)[5]  =                                                  VVV
//            pos0        pos1        pos2       pos3      pos4    pos5    pos6

//which is  9
 
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Expert Comment

by:fatimao
Comment Utility
hi!

The name of ur array i.e. currently buf, always point to the start of array (logical address) and as u might know that arrays are consecutie memory location.
So if we say that buf is pointing to memory area 1245 (buf -> 1245 ) then there are 10 locations reserved for it (as the space specified was 10) from location 1245 to 1254.

buf[0] -> 1  (memory location 1245)
buf[1] -> 2  (memory location 1246)
   .
   .
   .
buf[9] -> 8  (memory location 1254)


When u write buf[5] it is equivalent that u are pointing five positions ahead from the start of array i.e 1249.
More over (buf+1) is equivalent to buf[1] so wat (buf+1)[5] eventually means fifth location starting from the first location (1246)of array buf. i.e 6th location (1250). thus giving u the value 9.

hope it helps.
regards
fa
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by:dennis_george
Comment Utility
If you got your problem solved then please close this question.....

Dennis
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