Unable to use long long number types?

Aziz · 03-28-2009, 03:19 PM

Hello guys

I am trying this simple program:

Code:

#include <iostream>
#include <math.h>
#include <ctime>

int main(int argc, char *argv[])
{
     cout << endl << argv[1] << endl
        << atol(argv[1]) << endl
        << strtoul(argv[1],NULL,36) << endl <<
        (unsigned long long) argv[1] << endl <<
        (long long) argv[1] << endl;

     return 0;
}

I want to be able to transfer the argv[1] (which is a long long type) into a variable, but I am not able to convert it to a number.

Here is the output when I test the program:

./program 1124555552555562554

Code:

1124555552555562554
541357626
4294967295
18446744073705356114
-4195502

How can I convert argv[1] into a long long type so I can use it like:

long long number = argv[1];

Thanks for any hints,

graemef · 03-28-2009, 04:15 PM

maybe try sscanf and the format %ull - for unsigned long long

Aziz · 03-28-2009, 05:09 PM

Quote:

Originally Posted by graemef

maybe try sscanf and the format %ull - for unsigned long long

Thanks for the answer. If I try this number:

9223372036854775807

sscanf(argv[1],"%ull",&number);

number becomes:

18446744069414584320

However, reading the sscanf man I think you meant "%llu" which worked perfectly.

Thanks alot for your help, very much appreciated.

graemef · 03-28-2009, 05:21 PM

Sorry for that error, I should never reply before my first coffee of the day. Glad you got it sorted out.

ta0kira · 03-28-2009, 05:22 PM

Why don't you use strtoull? As far as I know, %llu is GNU but strtoull is C99 and POSIX. You'll at least have build problems to indicate lack of support for strtoull, but that won't happen with %llu (if you ever plan to use a different OS.)
Kevin Barry

PS In both cases you should be checking for residual data. For example, try entering 1234hello as your first argument and see if it's accepted. My guess is that it is, which should be undesirable.