printf %n question

The code:

#include <cstdio>


int main()
{
using namespace std;

int i;

printf("%ld\n%n", 1, &i);

printf("==>%d\n", i);

printf("%ld\n%n", 123456, &i);

printf("==>%d\n", i);

}

in my system produces:

[john@localhost src]$ ./foobar-cpp
1
==>2
123456
==>7

[john@localhost src]$


I expected:

"[john@localhost src]$ ./foobar-cpp
1
==>1
123456
==>6
[john@localhost src]$ "



It looks like it increments the value written to i with 1 more, than it
should. Is this a bug of my compiler, or am I wrong?

On 31 Oct, 09:42, john <j...@no.spamwrote:Interesting - have never come across %n before:
Googling unearths (in http://www.cplusplus.com) :
"Nothing printed. The argument must be a pointer to a signed int,
where the number of characters written so far is stored."

So your sample is behaving correctly:
"1" + "\n" = 2 characters
"123456" + "\n" = 7 characters
as reported.

On Wed, 31 Oct 2007 11:42:36 +0200, john <john@no.spamwrote in
comp.lang.c++:
You are wrong. You put the %n after the newline character in the
format string, so it is counted as well. Try putting the %n before
the \n.

--
Jack Klein
Home: http://JK-Technology.Com
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tragomaskhalos <dave.du.vergier@logicacmg.comwrites:
Very useful for IOCCC entries.

Chip

--
Charles M. "Chip" Coldwell
"Turn on, log in, tune out"
Somerville, Massachusetts, New England