Newbie data types question in printf vs iostream

[Mark Bruno]

When I declare a 64 bit integer (long long or __int64), it seems to have the
same boundaries as a 32 bit int while using printf, but the correct 64 bit
boundaries when using cout. Here's my code:

#include <cstdio>
#include <iostream>
using namespace std;
int main()
{
long long a = 2147483647; //The max value for a signed 32 bit int
long long b = 9223372036854775807; //The max value for a signed 64 bit int
printf("%d %d %d\n", a, a+1, b);
cout << a << " " << a+1 << " " << b << endl;
return 0;
}

 

[Arthur Gold]

When I declare a 64 bit integer (long long or __int64), it seems to have the
same boundaries as a 32 bit int while using printf, but the correct 64 bit
boundaries when using cout. Here's my code:

#include <cstdio>
#include <iostream>
using namespace std;
int main()
{
long long a = 2147483647; //The max value for a signed 32 bit int
long long b = 9223372036854775807; //The max value for a signed 64 bit int
printf("%d %d %d\n", a, a+1, b);
cout << a << " " << a+1 << " " << b << endl;
return 0;
}

Because that's what you're *telling* it to do; `%d' is for printing
`int'. The format specifier you want is `%lld' for long long.

[The C++ stream operators get it `right' because of overloading.]

HTH,
--ag

 

[Steven Green]

When I declare a 64 bit integer (long long or __int64), it seems to have the
same boundaries as a 32 bit int while using printf, but the correct 64 bit
boundaries when using cout. Here's my code:

#include <cstdio>
#include <iostream>
using namespace std;
int main()
{
long long a = 2147483647; //The max value for a signed 32 bit int
long long b = 9223372036854775807; //The max value for a signed 64 bit int
printf("%d %d %d\n", a, a+1, b);
cout << a << " " << a+1 << " " << b << endl;
return 0;
}

A 64bit noob here.
Is long long standard or is it really just long.

For g++ the compiler will treat long as a 64bit int if compiled with
the -m64 option (although my compiler hasn't been compiled to support it)

I would guess that the printf function doesn't know anything about a
long long and is merely treating it as though it were a 32bit int.

 

[Arthur Gold]

A 64bit noob here.
Is long long standard or is it really just long.

Good catch.

The type `long long' became standard in C99; it is not part of standard
C++ (though it's likely to become the case at some point). In any event,
it's an extension that's frequently available.

For g++ the compiler will treat long as a 64bit int if compiled with
the -m64 option (although my compiler hasn't been compiled to support it)

I would guess that the printf function doesn't know anything about a
long long and is merely treating it as though it were a 32bit int.

See my reply elsethread.

HTH,
--ag

 

[Mark Bruno]

D'OH!!! I forgot to use %llu!!! I'm so stupid! Thanks for pointing out my
error.

 

[Jack Klein]

A 64bit noob here.
Is long long standard or is it really just long.

In C, yes. In C++, no.

--
Jack Klein
Home: http://JK-Technology.Com
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