Something About Date Type

[Donkey]

Hello, I want to find out how many digits does each date type have and
how many bytes does it occupy in memory, so I wrote a program below:

#include <stdio.h>
const long double num=1123222.232 121342;
main()
{
Printf("the number occupies: %i, and it is 1123222.2321213 42
\n",sizeof num);
printf("char: %i ,the number is %c\n",sizeof(ch ar),(char)num);
printf("widecha r: %i\n",sizeof(wc har_t));
printf("singed char: %i\n",sizeof(si gned char));
printf("unsigne d char: %i\n",sizeof(un signed char));
printf("int: %i, the number is %i\n",sizeof(in t),(int)num);
printf("long int: %i, the number is %li\n",sizeof(l ong int),(long
int)num);
printf("float: %i, the number is %f\n", sizeof(float),f loat(num));
printf("double: %i, the number is
%lf\n",sizeof(d ouble),(double) num);
printf("long double: %i,the number is %Lf\n",sizeof(l ong
double),(long double)num);
printf("long long int: %i\n",sizeof(lo ng long int));
printf("long long double: %i\n",sizeof(lo ng long double));
}

But the execution result of the program is very surprising:

the number occupies: 12, and it is 1123222.2321213 42
char: 1 ,the number is 
widechar: 2
singed char: 1
unsigned char: 1
int: 4, the number is 1123222
long int: 4, the number is 1123222
float: 4, the number is 1123222.250000
double: 8, the number is 1123222.232121
long double: 12,the number is -0.000000
long long int: 8
long long double: 8

You may notice that:
the memory occupied by int Date type and long int Date type is the
same,
print of long double number is not correct ,
and the memory long long double date type occupied is smaller (8 bytes)
than long double date type(12 bytes).
The Compiler I Use is GCC. The OS I use is Windows XP.
Can anyone tells me why do these problems happen?
Thanks a million.

[Jonathan Burd]

Donkey wrote:

Hello, I want to find out how many digits does each date type have and
how many bytes does it occupy in memory, so I wrote a program below:

Which ``date'' type are you talking about?
#include <stdio.h>
const long double num=1123222.232 121342;
main()
int main (void) or
int main (int argc, char *argv[]) or
int main (int argc, char **argv).

Please read the newsgroup archive before posting code.
{
Printf("the number occupies: %i, and it is 1123222.2321213 42
\n",sizeof num);
Identifiers in C are case-sensitive. ``Printf'' is not the same as
``printf''
printf("char: %i ,the number is %c\n",sizeof(ch ar),(char)num);
Loss of data. Why print a double value as a char?
sizeof evaluates to a value of type size_t. You can use
a cast to ``unsigned long'' here and %lu for the conversion
specifier.
printf("widecha r: %i\n",sizeof(wc har_t));
printf("singed char: %i\n",sizeof(si gned char));
printf("unsigne d char: %i\n",sizeof(un signed char));
printf("int: %i, the number is %i\n",sizeof(in t),(int)num);
printf("long int: %i, the number is %li\n",sizeof(l ong int),(long
int)num);
printf("float: %i, the number is %f\n", sizeof(float),f loat(num));
printf("double: %i, the number is
%lf\n",sizeof(d ouble),(double) num);
printf("long double: %i,the number is %Lf\n",sizeof(l ong
double),(long double)num);
printf("long long int: %i\n",sizeof(lo ng long int));
printf("long long double: %i\n",sizeof(lo ng long double));
}
main returns int.

But the execution result of the program is very surprising:
Not really surprising.

<snip>

You may notice that:
the memory occupied by int Date type and long int Date type is the
same,
Again, which ``Date'' type? I don't see any such type in your code.

<snip>
Can anyone tells me why do these problems happen?
Thanks a million.

They happen because you forget to read a good book on C.
Pick up K&R 2. :)
Regards,
Jonathan.

--
Meow!

[Keith Thompson]

"Donkey" <ne********@hot mail.com> writes:

Hello, I want to find out how many digits does each date type have and
how many bytes does it occupy in memory, so I wrote a program below:

#include <stdio.h>
const long double num=1123222.232 121342;
main()
This should be "int main(void)".
{
Printf("the number occupies: %i, and it is 1123222.2321213 42\n",sizeof num);
printf, not Printf.

The "%i" format expects an int. sizeof yields a size_t.

To print the result of sizeof, convert it to unsigned long -- or use
"%zu" if your runtime library supports it (it's new in C99). For
example:

printf("sizeof num = %ld\n", (unsigned long)sizeof num);
printf("char: %i ,the number is %c\n",sizeof(ch ar),(char)num);
printf("widecha r: %i\n",sizeof(wc har_t));
printf("singed char: %i\n",sizeof(si gned char));
printf("unsigne d char: %i\n",sizeof(un signed char));
printf("int: %i, the number is %i\n",sizeof(in t),(int)num);
printf("long int: %i, the number is %li\n",sizeof(l ong int),(long int)num);
printf("float: %i, the number is %f\n", sizeof(float),f loat(num));
float(num) is a syntax error; you mean (float)num.
printf("double: %i, the number is %lf\n",sizeof(d ouble),(double) num);
printf("long double: %i,the number is %Lf\n",sizeof(l ong double),(long double)num);
printf("long long int: %i\n",sizeof(lo ng long int));
printf("long long double: %i\n",sizeof(lo ng long double));
There is no "long long double" in standard C.
}
The code you posted isn't the code you compiled (unless your compiler
is entirely too forgiving). It would also be helpful if you kept your
line length down.
But the execution result of the program is very surprising:

the number occupies: 12, and it is 1123222.2321213 42
char: 1 ,the number is 
widechar: 2
singed char: 1
unsigned char: 1
int: 4, the number is 1123222
long int: 4, the number is 1123222
float: 4, the number is 1123222.250000
double: 8, the number is 1123222.232121
long double: 12,the number is -0.000000
long long int: 8
long long double: 8

You may notice that:
the memory occupied by int Date type and long int Date type is the
same,
print of long double number is not correct ,
and the memory long long double date type occupied is smaller (8 bytes)
than long double date type(12 bytes).
The Compiler I Use is GCC. The OS I use is Windows XP.

It's not uncommon for types int and long int to be the same size.

Possibly your runtime library (which is separate from your compiler)
doesn't handle "%Lf" properly.

Recent versions of gcc properly reject "long long double". A buggy
old version treats it as "long long".

--
Keith Thompson (The_Other_Keit h) ks***@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.

[Keith Thompson]

Jonathan Burd <[e-mail] m> writes:

Donkey wrote:

Hello, I want to find out how many digits does each date type have and
how many bytes does it occupy in memory, so I wrote a program below:

Which ``date'' type are you talking about?

Perhaps it was a typo for "data type".

--
Keith Thompson (The_Other_Keit h) ks***@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.

[Peter Shaggy Haywood]

Groovy hepcat Keith Thompson was jivin' on Fri, 31 Mar 2006 06:17:34
GMT in comp.lang.c.
Re: Something About Date Type's a cool scene! Dig it!

To print the result of sizeof, convert it to unsigned long -- or use
"%zu" if your runtime library supports it (it's new in C99). For
example:

printf("sizeof num = %ld\n", (unsigned long)sizeof num);

No doubt Keith means this. (Note the %lu conversion specifier):

printf("sizeof num = %lu\n", (unsigned long)sizeof num);

--

Dig the even newer still, yet more improved, sig!

http://alphalink.com.au/~phaywood/
"Ain't I'm a dog?" - Ronny Self, Ain't I'm a Dog, written by G. Sherry & W. Walker.
I know it's not "technicall y correct" English; but since when was rock & roll "technicall y correct"?

[Keith Thompson]

ph******@alphal ink.com.au.NO.S PAM (Peter "Shaggy" Haywood) writes:

Groovy hepcat Keith Thompson was jivin' on Fri, 31 Mar 2006 06:17:34
GMT in comp.lang.c.
Re: Something About Date Type's a cool scene! Dig it!

To print the result of sizeof, convert it to unsigned long -- or use
"%zu" if your runtime library supports it (it's new in C99). For
example:

printf("sizeof num = %ld\n", (unsigned long)sizeof num);

No doubt Keith means this. (Note the %lu conversion specifier):

printf("sizeof num = %lu\n", (unsigned long)sizeof num);

Yes, thanks for the correction.

--
Keith Thompson (The_Other_Keit h) ks***@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.