uint64_t

[CMarco]

Hello to all,

I have a uint64_t variable and when I do:

uint64_t it = 100;

char msgname[2000];
sprintf(msgname, "Transport:%d+%d-", it, outGateBaseId+i);

It appears to have a format error because of uint64_t. Instead of %d I have already tried %i, %ll but without success.

Can someone advise me.

Regards
CMarco

[TheIdeasMan]

Try %u for unsigned int, not sure if works with long long, which is what uint64_t is (if not, just long)

[viliml]

Use stringstreams. Then you won't have to worry about that kind of stuff anymore, at all.

uint64_t it = 100;
stringstream msgname;
msgname<<"Transport:"<<it<<'+'<<outGateBaseId+i;

[HiteshVaghani1]

uint64_t i = 5;
printf ( "% PRIu64 ", i ) ;

and for that you have to include
#include <inttypes.h>

[TheIdeasMan]

I wasn't sure whether the OP's code had to be C.

[Peter87]

Have you tried %lu or %llu?

Or maybe this will work:
sprintf(msgname, "Transport:%"PRIu64"+%d-", it, outGateBaseId+i);

[viliml]

Or simply use stringstreams and forget about all the trouble!

[CMarco]

Hello

Thanks to all for the tips.

With the include and %llu the problem was solved.
The stringstreams solutions solve this also.

Regards

[vlad from moscow]

In C++ type name uint64_t is declared in header <cstdint> and is usually a typedef for unsigned long long. So you can use format specifier ll
From the C Standard

ll (ell-ell) Specifies that a following d, i, o, u, x, or X conversion specifier applies to a long long int or unsigned long long int argument; or that a
following n conversion specifier applies to a pointer to a long long int
argument.

[Cubbi]

uint64_t is unsigned long on one of the production systems at my work. PRIu64 is the right solution, for C.