int and char

[vgoel38]

#include<stdio.h>
#include<conio.h>

main()
{
char c='amsdlsn';
printf("%c",c);
getch();
}

output:
n

Q.1why is it that whatever i take the value of c, it has to be the last letter of the string?
Q.2the code works even if i replace char by int and %c by %d.why is it so?

[fg109]

1. I don't understand your question.
2. All characters have an integer value.
http://www.asciitable.com/

[vgoel38]

Q.1 see my ques is
if char c='sjfbksj' output is j
if cahr c='azhsb' output is b
the last letter is always the output..y is it so?

Q.2 so u mean int c='sgdrg' is a valid expression?
Also
if int c='a' output is 97
if int c='ab' output is 24930
why is it so?

[guestgulkan]

vgoel38 wrote:

Q.1 see my ques is
if char c='sjfbksj' output is j
if cahr c='azhsb' output is b
the last letter is always the output..y is it so?

Because a char can only hold 1 char obviously - so if you
try to give it a more than one char like you are trying to do - you will
only get the last one.

Q.2 so u mean int c='sgdrg' is a valid expression?
Also
if int c='a' output is 97
if int c='ab' output is 24930
why is it so?

in your case an int is 4 bytes long.
so if you say
int c = 'ABCDEFG' - then the compiler will take the last 4
chars and put them into the integer.
So c will be (in hex) 0x44454647 - which is (decimal) 1145390663

BUT the compiler will give you a warning if you try to fit too big a value
into a particular variable. The warning will be something like:
warning: character constant too long for its type


Can you see what is going on here???

[vgoel38]

q1..yeah but why is it that last digit is output and not any other?
q2..i coudn't understand following line of yours

then the compiler will take the last 4 chars
So c will be (in hex) 0x44454647