Displaying a 'long long' with printf

Hi all - I am having some difficulty displaying a variable of type 'long long' with printf(). My code is as follows:

Code:

---

/* File: largenum.c */
#include <stdio.h>

int main(int argc, char *argv[])
{
        // note that the last two characters are lowercase 'L', not '1'
        long long num = 317584931803ll;
       
        printf("Very Large Number: %lld \n", num );
        printf("Sizeof (long long) = %d\n", sizeof(long long) );
        return 0;
}

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This compiles with no warnings for me (mingw gcc 3.4.2). However, the output I recieve is always this:

Quote:

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Very Large Number: -242648101
Sizeof (long long) = 8

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If I convert the project to c++ and use cout to display num, it shows exactly the number I want it to, but I already am quite familiar with C++, and I want to learn how to do this in C.
The size of a long long is 8 bytes (as it should be), so I am fairly certain that the problem lies in my usage of printf(). If I convert it to an 'unsigned long long', and printf("%llu ", num) -- , I at least get a POSITIVE number, but it is still not the correct number. Any help would be appreciated.

I prefer to use the definitions from stdint.h:

Code:

---

itsme@dreams:~/C$ cat bigone2.c
#include <stdio.h>
#include <stdint.h>

int main(void)
{
  uint64_t num = UINT64_MAX;
  int64_t num2 = 317584931803ll;

  printf("%llu\n", num);
  printf("%lld\n", num2);

  return 0;
}

---

Code:

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itsme@dreams:~/C$ ./bigone2
18446744073709551615
317584931803

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That did the trick! Thanks alot. I guess, from reading this link, that I needed to declare it as 'long long int '.

Quote:

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Originally Posted by trinitrotoluene

Hi all - I am having some difficulty displaying a variable of type 'long long' with printf().

Code:

---

/* File: largenum.c */
#include <stdio.h>

int main(int argc, char *argv[])
{
        // note that the last two characters are lowercase 'L', not '1'
        long long num = 317584931803ll;
       
        printf("Very Large Number: %lld \n", num );
        printf("Sizeof (long long) = %d\n", sizeof(long long) );
        return 0;
}

---

This compiles with no warnings for me (mingw gcc 3.4.2).

---

Be sure that you are compiling in C99. Add -std=c99 on the command line.

[EDIT] nm [/EDIT]

gg

ok ok ok ok

Quote:

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Originally Posted by Emmanuel Delaha

Be sure that you are compiling in C99. Add -std=c99 on the command line.

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Is there a different approach to using this technique for really long numbers without using C99?

C99 is a standard that your compiler will either support, perhaps partially, or not at all. Otherwise the support for a 64bit integer will be a compiler specific feature. For example, MSVC supports 64bit integers using the (compiler-specific) built-in type "__int64".

...

I decided to post my original response - although it's not the root cause of trinitrotoluene's problems, it is a problem that's popped up here a few times in the past. In short, you can use the MS C-runtime library (MSVCRT.DLL) with MinGW - and the version of printf() in this library uses "%I64" instead of "%ll" when specifying a 64bit integer.
http://www.mingw.org/MinGWiki/index....8634fcd27af45d

gg

Quote:

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and the version of printf() in this library uses "%I64" instead of "%ll" when specifying a 64bit integer.

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You're absolutely correct! Sorry, itsme86, I jumped the gun when I thought that your solution fixed the problem. It didn't :( . I was still getting weird output like

Quote:

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4294967295
-242648101

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but changing the printf statements to "%I64u" and "%I64d" respectively (as suggested by Codeplug) produced the correct output. Since itsme86 was obviously using a *nix system (or maybe cygwin/msys, but I doubt it), and his code showed the output fine, does that mean that my code with the "%I64u" is now win32 platform specific? If I specify that I want ANSI compliance with the flag -pedantic-errors , I then get two warnings of

Quote:

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warning: ISO C does not support the `I64' printf length modifier

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Arg. So, is there an ANSI compliant way to display 64 bit integers with printf() ?

>> does that mean that my code with the "%I64u" is now win32 platform specific?
Yes.

>> is there an ANSI compliant way to display 64 bit integers with printf()?
When using the MS C-runtime (MSVCRT.DLL) - no.

The only alternative C-runtime for MinGW that I know of is within the Cygwin enviroment (using "newlib").

gg

Quote:

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Originally Posted by Kleid-0

Is there a different approach to using this technique for really long numbers without using C99?

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C90 only supports long and unsigned long.

If you are using an extension, the details belong to the compiler's documentation.