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Guess i was right according to King Mir's Explanation...

but the statement does NOT flag an error when the second operand is not enclosed within parenthesis.. ie., a<10?b=100:(b =200); // for a...

i am using Fedora core 5 gcc 4.1

Guess i ll have to read up on the gcc...

thanks for help everybody...

thanks.. but... here's one more doubt...

say a=0

a<10?b=100:b=200;

how come this statement works n "b" gets initialized to 100... in this case parenthesis is not required. i tried it out......

i am not completely convinced with it..
two doubts

1. If i am not mistaken the address of "b" is returned & is attempted to be initialized to 200.
But this error doesnt happen if the staement...

hi everybody..

i dont know whether the question i posted was clear enough or not... :confused: my question still remains unanswered... Let me restate the question..

I am AWARE that
...

#include<stdio.h>

int main()
{
int a=10,b=0;
a<0?b=100:b=200;
return 0;
}

sorry.. didnt realize that it got published...
will take care next time while posting..

sorry once again.. :(

#include<stdio.h>

int main()
{
float a=0.7;

if( a<0.7) //
printf("true\n");
else
printf("false\n");

hi itsme86...

i guess i wasnt clear when describing the functionality... (i have been fiddling with bit manipulation only from the past 2-3 days)

i guess it gave the impression that LSB and MSB...

for the new pbox:
input: 0xFE
output: 0xFD

hi ZuK...

i dont want to just swap the LSB & MSB...
The "pbox" can contain any "permutation" of the original bits..

regards

hi everyone...
i have pasted a piece of code below.. can u people please give me a review of this code... i need to know if there's a better way/logic to implement the following functionality.

...