Re: Bamboozled by long long

Re: Bamboozled by long long

[arijitg]

I tried running it on my system with both options.
 llx works while LX does not.

My system is sunOS5.7 just in case.

thanks guys for the prompt and right response. :)

---------Included Message----------
>Date: Fri, 19 Jul 2002 12:31:23 -0700
>From: "Gokhan Kisacikoglu" <kisa@centropolisfx.com>
>Reply-To: <kisa@centropolisfx.com>
>To: "John Love-Jensen" <eljay@adobe.com>
>Cc: <arijitg@uci.edu>, <gcc-help@gcc.gnu.org>
>Subject: Re: Bamboozled by long long
>
>
>> unsigned long long k=1;
>> unsigned int i = 4;
>> printf("The values are i: %d, , k: %LX, i+k: %LX \n", i,k , i+k);
>> k = k + i;
>> printf("The values are i: %d, , k: %LX, i+k: %LX \n", i,k , i+k);
>> 
>> ...depending on the extension (or convention for long long) used.
>> 
>> --Eljay
>
>
>This is not correct, this will only work with double precision float
>numbers. "long long" is really "long long int" and only ll should be
>used. This is from the man page (on my system):
>
>        ll   For n, the argument has type pointer to long long int; 
for
>d
>             and i, long long int; and for o, u, x, and X, unsigned 
long
>             long int.
>
>        L    For b, B, e, E, f, g, and G, the argument has type long
>             double.
>
>
---------End of Included Message----------

Bamboozled by long long

[arijitg]

 I have the following 2 lines in a program

 unsigned long long k=1;
 unsigned int i = 4;

 printf("The values are i: %d, , k: %lX, i+k: %lX \n", i,k , i+k);
 k = k + i;
 printf("The values are i: %d, , k: %lX, i+k: %lX \n", i,k , i+k);

I get the foll. outputs:

The values are i: 4, , k: 0, i+k: 1
The values are i: 4, , k: 0, i+k: 5

Note that (i+k) is showing k's current value, yet k is always showing 
0.

Any idea please?

Re: Bamboozled by long long

[John Love-Jensen]

Your data types are not matching your formats, a %lX expects a long, not a
long long.

Do this:

unsigned long long k=1;
unsigned int i = 4;
printf("The values are i: %d, , k: %lX, i+k: %lX \n", i, (long)k,
       (long)(i+k));
k = k + i;
printf("The values are i: %d, , k: %lX, i+k: %lX \n", i, (long)k,
       (long)(i+k));

That'll fix the problem.  If you have an appropriately modified/enhanced
"standard" C library, you may be able to do this...

unsigned long long k=1;
unsigned int i = 4;
printf("The values are i: %d, , k: %llX, i+k: %llX \n", i,k , i+k);
k = k + i;
printf("The values are i: %d, , k: %llX, i+k: %llX \n", i,k , i+k);

...or this...

unsigned long long k=1;
unsigned int i = 4;
printf("The values are i: %d, , k: %LX, i+k: %LX \n", i,k , i+k);
k = k + i;
printf("The values are i: %d, , k: %LX, i+k: %LX \n", i,k , i+k);

...depending on the extension (or convention for long long) used.

--Eljay

Re: Bamboozled by long long

[Gokhan Kisacikoglu]

> unsigned long long k=1;
> unsigned int i = 4;
> printf("The values are i: %d, , k: %LX, i+k: %LX \n", i,k , i+k);
> k = k + i;
> printf("The values are i: %d, , k: %LX, i+k: %LX \n", i,k , i+k);
> 
> ...depending on the extension (or convention for long long) used.
> 
> --Eljay


This is not correct, this will only work with double precision float
numbers. "long long" is really "long long int" and only ll should be
used. This is from the man page (on my system):

        ll   For n, the argument has type pointer to long long int; for
d
             and i, long long int; and for o, u, x, and X, unsigned long
             long int.

        L    For b, B, e, E, f, g, and G, the argument has type long
             double.

Re: Bamboozled by long long

[Leo Przybylski]

Doesn't make the code very portable does it?

-Leo

On Fri, 2002-07-19 at 12:14, John Love-Jensen wrote:
> Your data types are not matching your formats, a %lX expects a long, not a
> long long.
> 
> Do this:
> 
> unsigned long long k=1;
> unsigned int i = 4;
> printf("The values are i: %d, , k: %lX, i+k: %lX \n", i, (long)k,
>        (long)(i+k));
> k = k + i;
> printf("The values are i: %d, , k: %lX, i+k: %lX \n", i, (long)k,
>        (long)(i+k));
> 
> That'll fix the problem.  If you have an appropriately modified/enhanced
> "standard" C library, you may be able to do this...
> 
> unsigned long long k=1;
> unsigned int i = 4;
> printf("The values are i: %d, , k: %llX, i+k: %llX \n", i,k , i+k);
> k = k + i;
> printf("The values are i: %d, , k: %llX, i+k: %llX \n", i,k , i+k);
> 
> ...or this...
> 
> unsigned long long k=1;
> unsigned int i = 4;
> printf("The values are i: %d, , k: %LX, i+k: %LX \n", i,k , i+k);
> k = k + i;
> printf("The values are i: %d, , k: %LX, i+k: %LX \n", i,k , i+k);
> 
> ...depending on the extension (or convention for long long) used.
> 
> --Eljay
>