Let's say u have an array of char called
char arrayofchar[30]

how do u display the content of the array at the 20th location.
I have tried this but it doesn't work.
printf ("%s", arrayofchar[20]);

You are telling printf() that you want to display a null-terminated
character string, when you really want to display a single character.

You should use "%c"

Also, the _first_ position is "arrayofchar[0]", the twentieth is
"arrayofchar[19]".

--
regards,

nick

The element at the 20th location is a character not a string so
you need "%c" as your conversion format.

If you don't think of a 0th element the the 20th element is at
index 19 rather than 20.

The ouput will probably not actually appear until you send a '\n'
character.

I think you want:
  printf("%c\n",arrayofchar[19]);

Malcolm Kay

printf("%c", arrayofchar[19]);        /* The 20th char */
printf("%s", arrayofchar+19); /* The string (if arrayofchar
                                   is 0-terminated) starting at the
                                   20th char */

--

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know how to learn. - Henry Adams

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aHmING a crit dans le message ...

Sure. You have an array of char, so an element is a char. The correct
formatter (?) for printf() is "%c".

printf ("%c\n", arrayofchar[20]);

and don't forget to include <stdio.h>

--
-hs- "Stove"
CLC-FAQ: http://www.eskimo.com/~scs/C-faq/top.html
ISO-C Library: http://www.dinkum.com/htm_cl
"Really?  When run on my machine, a printed copy of the C FAQ leaps
from the monitor and whacks me over the head.." -- Chris Mears CLC

and of course, the 20th location is at index 19.

printf ("%c\n", arrayofchar[19]);

--
-hs- "Stove"
CLC-FAQ: http://www.eskimo.com/~scs/C-faq/top.html
ISO-C Library: http://www.dinkum.com/htm_cl
"Really?  When run on my machine, a printed copy of the C FAQ leaps
from the monitor and whacks me over the head.." -- Chris Mears CLC

Thanks guys!!!