You can just cast the value to an int (in both the boost::format example, and in the iostreams example) to get what you want. (the varargs usage you demonstrate with printf is essentially a much more dangerous form of casting anyway). -- Brian On Dec 18, 2006, at 10:13 AM, Phil Endecott wrote:

Dear All,

This does what I want:

uint8_t i = 65; printf("i = %d\n",i);

It prints "65". This doesn't:

uint8_t i = 65; cout << "i = " << i << "\n";

It treats i as a char, since uint8_t is a typedef for char deep in some header file somewhere, and prints "A".

I was hoping that boost::format would do what I wanted:

uint8_t i = 65; cout << boost::format("i = %d\n") % i;

But it seems to have the same, unwanted, behaviour as cout<

Is there anything that I can do to my code to get the effect that I want? If not, is there anything that can be changed in boost::format to make it do what I want? At the moment I have to write things like

cout << "i = " << static_cast<unsigned int>(i) << "\n";

which is a lot of typing and obfuscates what is actually going on.

Any suggestions would be much appreciated.

Cheers,

Phil.

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