boost::format("%p") % (uint8_t *) str is not like printf("%p", str)

[Geoff Barrett <gbarrett@…>]

I'm actually on 1.33.1 still but I can't see any difference that fixes this in the latest code.

The boost::format gets pointers to 8-bit types wrong. This because the underlying type is indistinguishable from a string. In this case, I think the formatting needs to take the hint from the format string about how to interpret the type.

I have a fix which is slightly hacky. Instead of passing the printf parameter directly to %, it can be wrapped in a container whose insertion method inspects the stream flags to see whether any of the basefield bits are set and only prints like a char/string if none of them are. The wrapper type is in the attachment. The format parsing needs to clear the basefield bits. Here is the diff from format/parsing.hpp:

321d320 < fpar->fmtstate_.flags_ &= ~std::ios_base::basefield; 326d324 < fpar->fmtstate_.flags_ &= ~std::ios_base::basefield;

[Marshall Clow]

This still a problem in 1.42:

#include <iostream>
#include <boost/format.hpp>
#include <cstdlib>

int main ( int argc, char *argv[] ) {
	const char *foo = "Hi Mom!";
	std::cout << str ( boost::format ( "%p" ) % foo ) << std::endl;
	printf ( "%p\n", foo );
	return 0;
	}

prints two different lines.

[Marshall Clow]

My copy of the C99 standard (n869) says about "%p":

p The argument shall be a pointer to void. The value of the pointer is converted to a sequence of printing characters, in an implementation-defined manner.

Given that, I don't see any way that Boost.Format can "match what printf prints".