Programming

I have a C test next week,so someone please help me in understanding the output of following 2 codes

1.

Code:

#define T t 
void main() 
{ 
char T = `T`; 
printf("
%c %c
",T,t); 
}

the output is T T, but since a macro is defined, i thought it would be t t


2.

Code:

void main() 
{ 
int a=0; 
int b=0; 
++a == 0 || ++b == 11; 
printf("
%d,%d",a,b); 
}

the output is 1 1 but the condition is false, so why this output?

For question number one:

T is substituted by t in the code-part so

becomes
and

becomes
Then I hope it becomes obvious.


For question two there is no real condition test to care about as I see it.
++a makes a = 1.. tested against the value 11 but the result ain't used anywhere.
Same for ++b == 11 as I see it.

The change scottn suggested makes no difference as I tested it either.

Guess the style of coding should be avoided if we have no idea what it's about

You're right, I made another change to the code. It didn't seem to make any sense after I wrote it, but before I could fix it, you replied. You were too quick

It seems you were expecting 'T' to get transformed into 't'. This doesn't happen because nothing inside a character constant or string constant is changed by the preprocessor.


The condition "a = 0; ++a == 0" is false - but remember the increment will still happen. Then, because the first condition is false, the second one needs to be tested to see if the whole statement is true or false. Whatever the truth value of the second part, the increment to b still happens because it's evaluated.