Re: printf("%d",float)

santosh <santosh.k83@xxxxxxxxx> writes:
candy_init@xxxxxxxxx wrote:
I just came across the following program:

#include <stdio.h>
int main()
{
float a = 12.5;
printf("%d\n", a);

printf expects and treats the corresponding argument as a signed int,
while you've passed it a float value.

printf("%d\n", *(int *)&a);

Same as above and an additional problem:
[...]

Not quite. The argument is of type int, so "%d" is the correct
format. The problem, of course, is that obtaining the value of that
argument (evaluating ``*(int *)&a'') invokes undefined behavior.

--
Keith Thompson (The_Other_Keith) kst-u@xxxxxxx <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."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
.

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