What format specifier, modifier, do I need for a 64 bit unsigned int
in [Learn C or C++]
|
From: Andrew Falanga on 27 May 2008 11:22 Hi, I know that %u is for unsigned integers, but, if memory serves me, that's only for 32 bit integers. What modifier would I use just before the 'u'? For example, (again if memory serves), to display an object of type long long, I would use "%ll", correct? So, what's necessary for an unsigned 64 bit integer? You know, I don't know if it would make any difference but I'm using a 64 bit compiler. I'm wondering if, in this case, it matters at all. Since the word size of the compiler is 64 bits, would I simply use %u and call it good? Andy
From: Richard Heathfield on 27 May 2008 11:31 Andrew Falanga said: > Hi, > > I know that %u is for unsigned integers, but, if memory serves me, > that's only for 32 bit integers. Not so. It is for unsigned ints. > What modifier would I use just > before the 'u'? For example, (again if memory serves), to display an > object of type long long, I would use "%ll", correct? Nearly. To display the /value/ of an object of type long long, in a decimal representation, you would use "%lld". > So, what's > necessary for an unsigned 64 bit integer? It depends on the type. If it's a short int, use "%hu". If it's an int, use "%u". If it's a long int, use "%lu". And if it's a long long int (always assuming your platform supports them, of course), use "%llu". C imposes no upper limit on the width of the basic integral data types. Even char is allowed to be 64 bits wide - or more! > You know, I don't know if it would make any difference but I'm using a > 64 bit compiler. C is C. It doesn't matter how many bits your compiler has, as long as it conforms to the language spec. > I'm wondering if, in this case, it matters at all. > Since the word size of the compiler is 64 bits, would I simply use %u > and call it good? No, your choice should be based on the specific type, not on bit-count. -- Richard Heathfield <http://www.cpax.org.uk> Email: -http://www. +rjh@ Google users: <http://www.cpax.org.uk/prg/writings/googly.php> "Usenet is a strange place" - dmr 29 July 1999
From: Andrew Falanga on 27 May 2008 11:45 On May 27, 9:31 am, Richard Heathfield <r...(a)see.sig.invalid> wrote: > Andrew Falanga said: > > > Hi, > > > I know that %u is for unsigned integers, but, if memory serves me, > > that's only for 32 bit integers. > > Not so. It is for unsigned ints. > > > What modifier would I use just > > before the 'u'? For example, (again if memory serves), to display an > > object of type long long, I would use "%ll", correct? > > Nearly. To display the /value/ of an object of type long long, in a decimal > representation, you would use "%lld". > > > So, what's > > necessary for an unsigned 64 bit integer? > > It depends on the type. If it's a short int, use "%hu". If it's an int, use > "%u". If it's a long int, use "%lu". And if it's a long long int (always > assuming your platform supports them, of course), use "%llu". C imposes no > upper limit on the width of the basic integral data types. Even char is > allowed to be 64 bits wide - or more! > > > You know, I don't know if it would make any difference but I'm using a > > 64 bit compiler. > > C is C. It doesn't matter how many bits your compiler has, as long as it > conforms to the language spec. > > > I'm wondering if, in this case, it matters at all. > > Since the word size of the compiler is 64 bits, would I simply use %u > > and call it good? > > No, your choice should be based on the specific type, not on bit-count. > > -- > Richard Heathfield <http://www.cpax.org.uk> > Email: -http://www. +rjh@ > Google users: <http://www.cpax.org.uk/prg/writings/googly.php> > "Usenet is a strange place" - dmr 29 July 1999 Thank you. To be honest, it's been a while since I've had to use format specifiers. Most of the code I've written lately I've been able to use the C++ I/O objects. Although this is C++, I'm coming into the project late and don't quite know the full extent of the environment. I don't know why, but we have to use format specifiers for output. Thanks again. Andy
From: Andrew Falanga on 28 May 2008 10:13 On May 27, 7:58 pm, Barry Schwarz <schwa...(a)dqel.com> wrote: > On Tue, 27 May 2008 08:45:29 -0700 (PDT), Andrew Falanga > > > > <af300...(a)gmail.com> wrote: > >On May 27, 9:31 am, Richard Heathfield <r...(a)see.sig.invalid> wrote: > >> Andrew Falanga said: > > >> > Hi, > > >> > I know that %u is for unsigned integers, but, if memory serves me, > >> > that's only for 32 bit integers. > > >> Not so. It is for unsigned ints. > > >> > What modifier would I use just > >> > before the 'u'? For example, (again if memory serves), to display an > >> > object of type long long, I would use "%ll", correct? > > >> Nearly. To display the /value/ of an object of type long long, in a decimal > >> representation, you would use "%lld". > > >> > So, what's > >> > necessary for an unsigned 64 bit integer? > > >> It depends on the type. If it's a short int, use "%hu". If it's an int, use > >> "%u". If it's a long int, use "%lu". And if it's a long long int (always > >> assuming your platform supports them, of course), use "%llu". C imposes no > >> upper limit on the width of the basic integral data types. Even char is > >> allowed to be 64 bits wide - or more! > > >> > You know, I don't know if it would make any difference but I'm using a > >> > 64 bit compiler. > > >> C is C. It doesn't matter how many bits your compiler has, as long as it > >> conforms to the language spec. > > >> > I'm wondering if, in this case, it matters at all. > >> > Since the word size of the compiler is 64 bits, would I simply use %u > >> > and call it good? > > >> No, your choice should be based on the specific type, not on bit-count. > > >> -- > >> Richard Heathfield <http://www.cpax.org.uk> > >> Email: -http://www. +rjh@ > >> Google users: <http://www.cpax.org.uk/prg/writings/googly.php> > >> "Usenet is a strange place" - dmr 29 July 1999 > > >Thank you. To be honest, it's been a while since I've had to use > >format specifiers. Most of the code I've written lately I've been > >able to use the C++ I/O objects. Although this is C++, I'm coming > >into the project late and don't quite know the full extent of the > >environment. I don't know why, but we have to use format specifiers > >for output. Thanks again. > > I have no idea if C++ adopted the exact width integer types of C99 but > if your system supports int64_t and uint64_t (the only types > guaranteed to be exactly 64 bits) then it should support the > corresponding formats (PRId64 and PRIu64). > > Remove del for email Thanks. I learned something interesting. Using %lu produced nothing but a '?' in the output as did using %u. The only thing that works is %ld, but this would be for a signed in, no? Andy
From: Andrew Falanga on 28 May 2008 17:15
On May 28, 10:33 am, Keith Thompson <ks...(a)mib.org> wrote: > Andrew Falanga <af300...(a)gmail.com> writes: > > [ questions and answers about printf formats deleted ] > > > Thanks. I learned something interesting. Using %lu produced nothing > > but a '?' in the output as did using %u. The only thing that works is > > %ld, but this would be for a signed in, no? > > "%ld" is for signed long int (also known as signed long, long int, or > long). "%lu" is for unsigned long int. > > If "%ld" works, I'd be very surprised if "%lu" produces just a '?'. > Can you show us some actual code that exhibits this behavior? > > Note: By "actual code" I mean a small, self-contained, compilable > program that we can try outselves without modification. > Copy-and-paste your actual code; don't re-type it. > > -- > Keith Thompson (The_Other_Keith) ks...(a)mib.org <http://www.ghoti.net/~kst> > Nokia > "We must do something. This is something. Therefore, we must do this." > -- Antony Jay and Jonathan Lynn, "Yes Minister" I wished I could give you what you're asking for here but in this case I can't. Besides, I'm so new to the build environment that, to be honest if I could copy/paste, it wouldn't do much good. Nowhere do I see people using printf() or other standard calls. It's all things like: Print() UINT64 and so forth. Since both of these, on the surface anyway, go outside of standard C or C++, it's beyond the scope anyway. I don't even know what include files are necessary for all this magic to take place. Basically, this is what I have (that would be compilable but for my lack of understanding the environment): // some includes here, don't know what int main() { UINT64 bigInt = 3; Print("bigInt contains %lu\n", bigInt); return 0; } That's what I have, but the output looks like: bigInt contains ? The compiler isn't catching that the %ld is for a signed, rather than unsigned, 64bit int. For this reason, I'm betting that somewhere, someone's not turning on something like "-Wformat" (that's for GNU gcc). I dug up that little tid-bit after finding out why I've been slapped before in the past for using a format specifier that didn't match the data type. Shoot, this environment is so unfamiliar to me that I've never seen a system where you need to define the entry point in a header file, but you do with this. Andy |