why the printf is not typecasting the int to float

[jayapal]

#include <stdio.h>

main ()
{
int k = 10;
printf ( "%f\n", k);

}
o/p ............... .... run time error

Why this program is showing run time error ... Is there any wrong in
these code part.

[jameskuyper]

jayapal wrote:

#include <stdio.h>

main ()
{
int k = 10;
printf ( "%f\n", k);

}
o/p ............... .... run time error

Why this program is showing run time error ... Is there any wrong in
these code part.

The printf() family uses the variadic function interface (see
<stdarg.h>), or at least one that is functionally equivalent to it.
When you call variadic functions, variable arguments are subject only
to the default argument promotions. The promoted type (in this case
'int') must match exactly the type specified in the format string
(which in this case is 'double'). You can explicitly convert k to
double, or you can change the format specifier to 'd', but one way or
another you have to make them match, or the behavior is undefined.

[Martin Ambuhl]

jayapal wrote:

>
#include <stdio.h>

main ()
{
int k = 10;
printf ( "%f\n", k);

^^
The compiler sees as arguments a string and an int. The usual promotion
rules for arguments call for no promotion.
The "%f" is part of a string interpreted by the printf function and is
meaningless as far as compilation is concerned except as chars in a string.
So the printf function is passed an int, is told that it is a double,
and promptly chokes, as can be expected.

If you want to printf the value of k as floating-point, you must make it
one:
printf("%f\n", (double)k);

[user923005]

On Dec 12, 11:12 am, jayapal <jayapal...@gma il.comwrote:

#include <stdio.h>

main ()
{
int k = 10;
printf ( "%f\n", k);

}

o/p ............... .... run time error

Why this program is showing run time error ... Is there any wrong in
these code part.

There is plenty wrong with it. This should do what you want. Look at
each difference between this program and your program. Each
difference is important.

#include <stdio.h>
int main(void)
{
const int k = 10;
printf("%f\n", (double) k);
return 0;
}

[Walter Roberson]

In article <a1************ *************** *******@q77g200 0hsh.googlegrou ps.com>,
user923005 <dc*****@connx. comwrote:

>On Dec 12, 11:12 am, jayapal <jayapal...@gma il.comwrote:

>#include <stdio.h>

> main ()
{
int k = 10;
printf ( "%f\n", k);

>}

>This should do what you want. Look at
each difference between this program and your program. Each
difference is important.

>#include <stdio.h>
int main(void)
{
const int k = 10;
printf("%f\n", (double) k);
return 0;
}

What is the "important" difference between using int k = 10 or
const int k = 10 ? Will the abstract behaviour of the program
be different in -any- way by using 'const' or not using 'const'
in that program? Is there, for example, a difference in the
behaviour of the cast to double? Why is it important that k be const
but that you do not cast k to (const double) ?

--
"Is there any thing whereof it may be said, See, this is new? It hath
been already of old time, which was before us." -- Ecclesiastes