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

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.

Sys_Prog

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

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dennis_george

refer the previous question
https://www.experts-exchange.com/questions/20774407/array-and-pointer-initilization.html

Dennis

Ajar

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

fatimao

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

dennis_george

If you got your problem solved then please close this question.....

Dennis