why does this prg does not display "NONE"

[sunny]

int a=10;
switch(a)
{
case '1':
printf("ONE\n") ;
break;
case '2':
printf("TWO\n") ;
break;
defa1ut:
printf("NONE\n" );
}

[Richard Heathfield]

sunny said:

int a=10;
switch(a)
{
case '1':
printf("ONE\n") ;
break;
case '2':
printf("TWO\n") ;
break;
defa1ut:

Well, duh.

printf("NONE\n" );
}

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

[Preben Traerup]

sunny wrote:

int a=10;
switch(a)
{
case '1':
printf("ONE\n") ;
break;
case '2':
printf("TWO\n") ;
break;
defa1ut:
printf("NONE\n" );
}

propably because a does not have value defalut :-)
../Preben

[ramasubramanian.rahul]

check the spelling of default ... and dont waste others time with such
hopeless questions
Preben Traerup wrote:

sunny wrote:

int a=10;
switch(a)
{
case '1':
printf("ONE\n") ;
break;
case '2':
printf("TWO\n") ;
break;
defa1ut:
printf("NONE\n" );
}

propably because a does not have value defalut :-)
./Preben

[zorigaman]

sunny a écrit :

int a=10;
switch(a)
{
case '1':
printf("ONE\n") ;
break;
case '2':
printf("TWO\n") ;
break;
default:
printf("NONE\n" );
}

Because it seems that with the quotes, you're evalutating the character
(ascii code), not the value itself.
Try :

switch(a)
{
case 1 :

[Clever Monkey]

sunny wrote:

int a=10;
switch(a)
{
case '1':
printf("ONE\n") ;
break;
case '2':
printf("TWO\n") ;
break;
defa1ut:
printf("NONE\n" );
}

Why do people continue to post this lame snippet?