Programmatically Ignore printf

Question

I'm using someones library that printf's out an error message when a connection to a device is unsuccessful. The library code prints out nothing when a connection is successful.

I periodically check (loop & sleep) to see if a device is connected but I only want to print out when it is connected.

At the moment I get something like:

Waiting for connection... (<-- My print)
Error
Error
Error
Error
Error
Connection successful (<-- My print)

What I want is:

Waiting for connection... (<-- My print)
Connection successful (<-- My print)

How can I programatically ignore a printf?

N.b. I found a similar question Programmatically Ignore Cout but that solution did not work for printf.

I am using Windows.

Can anyone help? (c/c++ novice)

I guess this question could fall under several tags c, c# maybe? Can someone edit if that is the case? (I've only had experience with c & c++) — Kevvvvyp, May 14 '15 at 10:30
@Kevin: No, tag the language you're actually using, not every language that supports console output. — Lightness Races in Orbit, May 14 '15 at 10:32
On POSIX, you can use `dup2` to redirect `printf`, and on most Unixes, that redirection can go nowhere via `/dev/null`. However, there is not a standard C++ way. — Potatoswatter, May 14 '15 at 10:44
@KevinPaton: At least [one answer there shows you how to do exactly that](http://stackoverflow.com/a/13817208/560648). — Lightness Races in Orbit, May 14 '15 at 10:46

W.F. · Accepted Answer · 2015-05-14T10:39:25.060

2

Have you tried something like (before establishing connection):

FILE * myout = stdout;
stdout = fopen ("standard-output-file", "w");

To print something to output you could use then:

fprintf(myout, "format", ...);

Edit: Remember to close the file descriptor afterwards:

fclose(myout);

edited May 14 '15 at 10:39

answered May 14 '15 at 10:36

W.F.

13,888
2
34
81

I am not wanting to print to a file (if I have understood you correctly), I am wanting to programatically ignore printfs temporarily. – Kevvvvyp May 14 '15 at 10:39
1

@KevinPaton Yes but you could open e.g. /dev/null as an output if you are in Linux that would be equivalent with ignoring it... – W.F. May 14 '15 at 10:40
@KevinPaton on the other hand everything written to `myout` would be shown in standard output as expected... – W.F. May 14 '15 at 10:41
It's implementation defined (not everywhere stdout is an assignable FILE*). For example (on Windows) current MSVC++ implementation changed to define stdout as return value of a function (actually function returns an array). – Adriano Repetti May 14 '15 at 11:00

Petr · Answer 2 · 2015-05-14T10:46:51.013

0

I was able to get this working under Linux the following way

#include <cstdio>

using namespace std;

int main() {
   printf("Start\n");
   freopen("/dev/null", "w", stdout);
   printf("middle\n");
   freopen("/dev/stdin", "w", stdout);  // really stdin here!
   printf("End\n");
   return 0;
}

Under Windows I think there are other names for files (nul and con maybe).

However, this seems to be completely unportable, and, what's worse, this will prevent the users from redirecting your program output to a file, because I explicitly reopen /dev/stdin. There are suggestions that you can use /dev/fd/1 instead of /dev/stdin to mitigate the latter problem, but I have not tested it.

It seems that there is no reliable way, see, e.g. http://c-faq.com/stdio/undofreopen.html

edited May 14 '15 at 10:46

answered May 14 '15 at 10:42

Petr

9,812
1
28
52

http://stackoverflow.com/a/13817208/153285 should be reliable, and about as portable as this. – Potatoswatter May 14 '15 at 10:48
You could also redirect stdio by wrapping the program executeable in a bash/batch script. – deek0146 May 14 '15 at 10:49
It's implementation defined (not everywhere stdout is an assignable FILE*). Moreover OP has to do it in Windows... – Adriano Repetti May 14 '15 at 10:59

score 0 · Answer 3 · answered May 14 '15 at 10:50

0

I don't know the absolute solution, but you'll need to use:

freopen("conout$", "w", stderr);

Or something similar, but with "conout$" pseudo-file, which I believe is same as /dev/stdout on Linux.

You may have to use GetStdHandle, AttachConsole etc.

answered May 14 '15 at 10:50

Ajay

18,086
12
59
105

score 0 · Answer 4 · answered May 14 '15 at 10:56

0

If you are using c++11 and above you can use a simple variadic template without much coding so this overloads printf for all cases and the original one never gets called

template <typename ...Args>
void printf(Args... ...)
{
}

answered May 14 '15 at 10:56

tejas

1,795
1
16
34

It won't work in 3rd party library (unless they're doing printf in their headers). – Adriano Repetti May 14 '15 at 10:59
unless he can recompile it. anyway I gave a shot. – tejas May 14 '15 at 11:03

Programmatically Ignore printf

4 Answers4