long long data type

[dijaster]

Recently I have had the need to run loops of at least 10^10 steps.
Since 32 bit integers apparently can only store values up to about
2*10^9, I have tried to use the "long long" data type. However, it has
not worked and I am unsure why. For example, I try the following code:

testlong.c:

#include <stdio.h>
#include <stdlib.h>

int main() {
long long a;
a = 10000000000LL; //10^10
printf("a = %i\n", a);
return EXIT_SUCCESS;
}

and compile it by:

gcc testlong.c -o testlong

then when I run the executable I obtain the following:

a = 1410065408

Does anyone know how to fix this problem? Perhaps gcc requires some
commandline argument to support long long ints?
Thanks for any suggestions.

[Robert Bauck Hamar]

Re: long long data type

dijaster wrote:[color=blue]
> Recently I have had the need to run loops of at least 10^10 steps.
> [...]. For example, I try the following code:
>
> testlong.c:
>
> #include <stdio.h>
> #include <stdlib.h>
>
> int main() {
> long long a;
> a = 10000000000LL; //10^10
> printf("a = %i\n", a);[/color]

As the format string specifies, printf expects it's second argument to
be an int.
[color=blue]
> return EXIT_SUCCESS;
> }
>
> and compile it by:
>
> gcc testlong.c -o testlong
>
> then when I run the executable I obtain the following:
>
> a = 1410065408[/color]

You need to specify the long long length modifier (ll) to printf:
printf("%lli\n" , a);

--
Robert Bauck Hamar

[dijaster]

Re: long long data type

Thanks Robert. That seems to have solved the problem.

[Keith Thompson]

Re: long long data type

"dijaster" <dijaster@yahoo .com> writes:[color=blue]
> Thanks Robert. That seems to have solved the problem.[/color]

Note that there may be systems on which the compiler supports
"long long", but the runtime library's implementation of printf()
doesn't (or supports it in a non-standard way).

The "long long" type was added to the language in the 1999 version of
the standard (C99), which is not yet widely implemented. Many pre-C99
compilers support it as an extension, but as I recall consensus on the
name of the type was reached before it was reached on the format
strings for printf.

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Erik de Castro Lopo]

Re: long long data type

dijaster wrote:[color=blue]
>
> Recently I have had the need to run loops of at least 10^10 steps.
> Since 32 bit integers apparently can only store values up to about
> 2*10^9, I have tried to use the "long long" data type. However, it has
> not worked and I am unsure why. For example, I try the following code:
>
> testlong.c:
>
> #include <stdio.h>
> #include <stdlib.h>
>
> int main() {
> long long a;
> a = 10000000000LL; //10^10
> printf("a = %i\n", a);[/color]

Thats the wrong conversion character. Try:

printf("a = %llu\n", a);

Erik
--
+-----------------------------------------------------------+
Erik de Castro Lopo nospam@mega-nerd.com (Yes it's valid)
+-----------------------------------------------------------+
Linux, the UNIX defragmentation tool.

[Lawrence Kirby]

Re: long long data type

On Mon, 20 Dec 2004 08:25:24 +1100, Erik de Castro Lopo wrote:
[color=blue]
> dijaster wrote:[color=green]
>>
>> Recently I have had the need to run loops of at least 10^10 steps.
>> Since 32 bit integers apparently can only store values up to about
>> 2*10^9, I have tried to use the "long long" data type. However, it has
>> not worked and I am unsure why. For example, I try the following code:
>>
>> testlong.c:
>>
>> #include <stdio.h>
>> #include <stdlib.h>
>>
>> int main() {
>> long long a;
>> a = 10000000000LL; //10^10
>> printf("a = %i\n", a);[/color]
>
> Thats the wrong conversion character. Try:
>
> printf("a = %llu\n", a);[/color]

%llu is for unsigned long long, for long long use %lld.

You CAN also use %lli. However %d forms are usually preferred over %i
because %d was the original, %i was only a later addition; it is possible
that some old compilers may not support %i, and %d is what C programmers
tend to be used to. d can be easier to read than i in some character
sets. Also remember that %d and %i do different things in scanf()
conversions, and %d is the closer match to the printf() behaviour.

Lawrence