Printf question

[anonymous]

Hi CLCers,
The program below prints some number and i dont
understand why it is behaving like this. I am not supplying
the necessary arguments to the printf function and
according to my understanding the program shiould throw
some error. Please clear my doubts.

#include<stdio.h>
int main()
{
printf("%d");
return 0;
}

Thanks in advance.

Sha

 

[Joona I Palaste]

(e-mail address removed) scribbled the following:

Hi CLCers,
The program below prints some number and i dont
understand why it is behaving like this. I am not supplying
the necessary arguments to the printf function and
according to my understanding the program shiould throw
some error. Please clear my doubts.

#include<stdio.h>
int main()
{
printf("%d");
return 0;
}

You are creating undefined behaviour. printf() expects an argument,
because of the format string, but you didn't supply one. This, by
itself, creates undefined behaviour, so anything might happen,
including printing random numbers. Most probably it's reading a
value from the top of the stack, but whatever value that is is
implementation-defined.

 

[pete]

(e-mail address removed) scribbled the following:


You are creating undefined behaviour. printf() expects an argument,
because of the format string, but you didn't supply one. This, by
itself, creates undefined behaviour, so anything might happen,
including printing random numbers. Most probably it's reading a
value from the top of the stack, but whatever value that is is
implementation-defined.

It's undefined.
It need not be documented and it need not be repeatable.

 

[Joona I Palaste]

It's undefined.
It need not be documented and it need not be repeatable.

That's why I said "you are creating undefined behaviour", and that's
also why I said "most probably" instead of "certainly".

 

[Christopher Benson-Manica]

(e-mail address removed) spoke thus:

int main()
{
printf("%d");
return 0;
}

Your program produces undefined behavior in two ways - one, by failing
to specify an argument for the %d conversion specifier (as others
noted), and two, by failing to supply a newline at the end of your
program's output. Invoking undefined behavior means that the
implementation is free to do whatever it wants, up to and including
perpetrating terrorist acts.

 

[Joona I Palaste]

(e-mail address removed) spoke thus:

Your program produces undefined behavior in two ways - one, by failing
to specify an argument for the %d conversion specifier (as others
noted), and two, by failing to supply a newline at the end of your
program's output. Invoking undefined behavior means that the
implementation is free to do whatever it wants, up to and including
perpetrating terrorist acts.

Failing to supply a newline at the end of the program's output causes
undefined behaviour? Chapter and verse, please?

 

[Christopher Benson-Manica]

Failing to supply a newline at the end of the program's output causes
undefined behaviour? Chapter and verse, please?

Um... can I quote from the book of Revelation? ;( Apparently I
(badly!) mis-recalled previous discussion on the topic. Thanks for
the cluon...

 

[pete]

That's why I said "you are creating undefined behaviour",
and that's also why I said "most probably" instead of "certainly".

There is nothing about the value at the top of the stack
(assuming a stack)
which suggests that it is an implementation defined value.

 

[pete]

Failing to supply a newline at the end of the program's output causes
undefined behaviour? Chapter and verse, please?

It's from the definition of "stream".

Whether or not the last line of a text stream needs to be
newline terminated is implementation defined.

If it isn't newline terminated, then it isn't a portable stream.
If it isn't a stream, then what is it ?

 

[Dan Pop]

(e-mail address removed) scribbled the following:


You are creating undefined behaviour. printf() expects an argument,
because of the format string, but you didn't supply one. This, by
itself, creates undefined behaviour, so anything might happen,
including printing random numbers. Most probably it's reading a
value from the top of the stack, but whatever value that is is
implementation-defined. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

^^^^^^^^^^^^^^^^^^^^^^
Nope, it ain't. It's *unspecified*, but this doesn't matter, because
undefined behaviour has *already* been invoked.

Be *very* careful when using terms like "implementation-defined", because
their meaning is precisely defined by the C standard (and the definition
is not always what one would intuitively expect).

Dan

 

[Dan Pop]

Failing to supply a newline at the end of the program's output causes
undefined behaviour? Chapter and verse, please?

2 A text stream is an ordered sequence of characters composed into
lines, each line consisting of zero or more characters
plus a terminating new-line character. Whether the last line
requires a terminating new-line character is implementation-defined.

So, if the implementation requires it, but the program doesn't supply it
you have undefined behaviour by lack of specification of behaviour.

Dan

 

[Peter Shaggy Haywood]

Groovy hepcat Christopher Benson-Manica was jivin' on Mon, 19 Jan 2004
14:54:48 +0000 (UTC) in comp.lang.c.
Re: Printf question's a cool scene! Dig it!

(e-mail address removed) spoke thus:


Your program produces undefined behavior in two ways - one, by failing

^^^
Would you believe three ways?

to specify an argument for the %d conversion specifier (as others
noted), and two, by failing to supply a newline at the end of your

Three, failing to provide a prototype for printf().

--

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