pai wrote On 05/01/06 14:12,:
Hi ,
Can any one tell me how this statement of printf is behaving . how
the last digit is printed
int a=2,b=4,c=7;
printf("%d",printf("%d %d:",a,b));
//answer to this was 2 4:3
The output ought to have been "2 4:4" -- either you've
mis-reported it, or your C implementation is broken.
printf("%d",printf("%d %d %d:",a,b,c));
//answer to this 2 4 7:6
This looks right.
Can any one explain this behavior ?
Yes, I can! And so can you, if you study it carefully.
Pretend you're the computer, mechanically executing the
first of the two puzzling statements:
First, you're confronted with a printf() call whose
basic form is `printf(format, argument);'. Before you
can actually execute the call, you need to evaluate the
format (easy: it's a string literal) and the second
argument. That second argument is ...
... a function call, namely, a call to printf(). So
you (in your persona as the computer) execute this printf()
call, which produces the output "2 4:". Also, printf()
returns a value: the number of characters that were output.
The call generated four characters, so the value is 4.
Back to the outer printf(), which now looks like
`printf("%d", 4);'. This call produces the output "4".
It also returns the number of characters of output (1),
but since you don't use that value it's simply ignored.
All together, the inner printf() outputs "2 4:" and
the outer printf() tacks on another "4", so the complete
output should be "2 4:4". Accept no substitutes.
Exercise for the reader: Analyze the second statement
the same way.
