Thread: Problem about float

I just checking but confused with float. in that code same size int,and same type double are working but float showing nothing in printf..why?? i'm using GCC compiler int 32bit win7 os

Code:

#‎include <stdio.h>
int main()
{

  char arr[10] = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' };

  printf("Size of char=%c\n", ((char *) (&arr[0]))[1]);

  printf("Size of short=%c\n", ((short *) (&arr[0]))[1]);

  printf("Size of int=%c\n", ((int *) (&arr[0]))[1]);

  printf("Size of long=%c\n", ((long *) (&arr[0]))[1]);

  printf("Size of float=%c\n", ((float *) (&arr[0]))[1]);

  printf("Size of double=%c\n", ((double *) (&arr[0]))[1]);

  return 0;
}

I'm surprised you get anything, since it's so obviously broken.

Code:

$ gcc foo.c
foo.c: In function ‘main’:
foo.c:13:2: warning: format ‘%c’ expects argument of type ‘int’, but argument 2 has type ‘long int’ [-Wformat]
foo.c:15:2: warning: format ‘%c’ expects argument of type ‘int’, but argument 2 has type ‘double’ [-Wformat]
foo.c:17:2: warning: format ‘%c’ expects argument of type ‘int’, but argument 2 has type ‘double’ [-Wformat]

The result of doing undefined stuff is NOT defined!!

Tim S.

i got output from those 6 printf like
1
2
4
4

8
as i expected but in 5th printf where float showing nothing..if float not showing then why double is working?? i checked that int* reached to 4bytes ahead as well as float went same address..but from that address %c should show 4.. why not showing by float*???? all are showing even same type double also but why not float*?

Originally Posted by Matticus

Refer to post #4 above (and post #2 also, for good measure).

What exactly are you trying to accomplish? Why not just use "sizeof"?

may be i failed to make you understand what i'm asking... i know about sizeof... i said in my post that i was just testing.....i know about those 3 warning,still i surprised..long and double are the holder of rest 2 warning but they are working as i told,but why not float??what wrong with that float* bit petterns...

If you really want to find the answer of what it is doing; then, find out how to have the compiler generate assembly output.
Then, read the assembly code.

Tim S.

All of them, except for char, would likely fail to produce the right answer on a big endian machine.

For example, the 'short' value would consist of a word containing the bytes '2' and '3', and it's only the quirk of the machine endianess that means it prints 2.

As for float, there is something else to consider.
Whenever a float is passed to a variadic function like printf, it is automatically promoted to double. You might think you were going to get lucky and just extract the LSB of the passed float, but what you actually get is the LSB of the promoted double. This of course has nothing to do with the original value you thought you had.

Originally Posted by Salem

All of them, except for char, would likely fail to produce the right answer on a big endian machine.

For example, the 'short' value would consist of a word containing the bytes '2' and '3', and it's only the quirk of the machine endianess that means it prints 2.

As for float, there is something else to consider.
Whenever a float is passed to a variadic function like printf, it is automatically promoted to double. You might think you were going to get lucky and just extract the LSB of the passed float, but what you actually get is the LSB of the promoted double. This of course has nothing to do with the original value you thought you had.

thanks.. i thought if it is automatically promoted to double then it may goes to at first of that address holds '8'. so if it is not gonna show 4 then it will show 8. now i got it, problem(to me) caused when it is converting to double .i checked it more and that made me more satisfied.. thanks.

i done it in other way.i saw that float* goes to my desired address but not printing next 8bits properly..so from that address i just do forward 8bits and again backed 8bits.then i tell to print next 8bits by %c ..it is working by printing 4

Code:

printf("Size of float=%c\n",((char*)(((float *) (arr))+1))[1]-1)

.. .i think i got it where the problem is

now without any warning my prog will work fine

Code:

#include<stdio.h>
int main()
{

  unsigned char arr[15] = { '0', '1', '2', '3', '4','5', '6', '7', '8', '9','a','b','c','d','e' };

  printf("Size of char=%c\n", ((char *) (&arr[0]))[1]);

  printf("Size of short=%c\n", ((short *) (&arr[0]))[1]);

  printf("Size of int=%c\n", ((int *) (&arr[0]))[1]);

  printf("Size of long=%c\n", ((char*)(((long *) (arr))+1))[1]-1);

  printf("Size of float=%c\n",((char*)(((float *) (arr))+1))[1]-1);

  printf("Size of double=%c\n", ((char*)(((double *) (arr))+1))[1]-1);

  return 0;
}