can anyone explain how printf works. is it's output implementation dependent? and also kindly explain the output for the following code.
# include <stdio.h>
void main()
{
int i=10;
printf ("%d %d %d",i,++i,++i);
}
i got
12 12 11
i don't understand how it works. kindly help.
thanx.
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darpangoel:
> can anyone explain how printf works. is it's output implementation dependent?
The page explains it best:
http://computer.howstuffwo
It has a section that covers printf exclusively...
> and also kindly explain the output for the following code.
imladris already covered it, so I'm not going to repeat it... :)
Hope That Helps,
Dex*
I'm not particularly certain why that program outputs 12, 12, 11, as I am a learning C coder myself, but I can explain a bit on printf.
The basic syntax of printf is: printf("<formatted output>", var1, var2, var3, var4);
Some of the switches I use routinely for printf are %d (integer), %e (scientific notation), %f (double, floating point), %c (character), and %s (string of characters).
The C compiler will basically go from left to right matching matching up fields with the variable arguments you give.
For instance, printf("%d<>%e<>%c", 3, 345.6, 'c'); would ouput : 3<>3.456000e+002<> c
Hope that helps a little with the primary question. I'm lost on finding an answer to the second one.
>>So, since i is 10, the first value pushed onto the stack will be 11 (10, preincremented), the second will be 12 (preincremented again), then 12 again, then the format specifier. And so it will print exactly what you saw: 12 12 11.
That makes perfect sense. Thanks for the clear explanation imladris.
http://msdn.microsoft.com/
"Function-call operator. All arguments to a function are evaluated and all side effects complete before entry to the function. No order of evaluation among the arguments is specified."
So, the order of evaluation of i, ++i and ++i is unspecified. The possible outcomes should be:
10, 11, 12 (1,2,3)
10, 12, 11 (1,3,2)
11, 11, 12 (2,1,3)
12, 11, 12 (2,3,1)
11, 12, 11 (3,1,2)
12, 12, 11 (3,2,1)
But not 12, 12, 12.
Imladris explained why 12, 12, 11 is the most likely order.
I would assume that the VC.NET compiler is in fact a C++ compiler. It may be that there are different rules for C++.
Also, this kind of thing (referring to i multiple times, some of which are pre (or post) incremented and some not) is often compiler dependant (i.e. left to the compiler implementors for decisions). So it is unwise to ever code such things in real life.
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by: imladrisPosted on 2003-11-14 at 11:02:20ID: 9750163
I'm not sure about your question about "output implementation dependant". However, the output is readily explainable.
In C, arguments are pushed onto the stack right to left. That is the rightmost argument goes onto the stack first, then the one next to it, etc. and the first argument is pushed last. This means that the called function finds the first argument at the "stack frame" (the spot the stack will be popped to when it needs to return).
So, since i is 10, the first value pushed onto the stack will be 11 (10, preincremented), the second will be 12 (preincremented again), then 12 again, then the format specifier. And so it will print exactly what you saw: 12 12 11.