problems with indirect width specification

[udit1992]

Hi,
I have some doubts related to indirect width specification.

Here is the sample code:

#include<stdio.h>
#include<conio.h>

main()
{
  float f=20, width_f=3;
  clrscr();
  printf("%*.2f",width_f,f);
  getch();
}

The output of the above code is 0.00; while if I change the data type of width_f from float to int, then the output comes out to be 20.00. I know that the default data type for width specification needs to be 'int'. But I am still not clear about the output when the data type is of float type.

Can anyone help me out with this ?? Thanks in advance.

[Salem]

Width specifications are always integer, regardless of the type you're trying to print.

[AndiPersti]

I know that the default data type for width specification needs to be 'int'. But I am still not clear about the output when the data type is of float type.

Based on your result and the use of conio.h I assume you are on a intel/windows machine where int is probably 4 bytes, float too, double is 8 bytes and numbers are stored little-endian. Additionally floating point numbers are represented in IEEE754 format. (correct me if I'm wrong).

First thing you have to know is that if you call a variadic function like printf (in other words a function which accepts a variable number of arguments), all float values will be promoted to double values.

Now I think the following is happening:
Your first value (width_f) is promoted to a double thus 3 is stored as 00 00 00 00 00 00 08 40 in memory. The second value (f = 20) is stored as 00 00 00 00 00 00 34 40. (You can use this online conversion tool)
The whole memory block looks like:
00 00 00 00 00 00 08 40 00 00 00 00 00 00 34 40

printf expects an int for the width format specifier which means it just reads the first 4 bytes which in your case are 00 00 00 00. Then it reads the next 8 bytes (%f means print a double) which are 00 00 08 40 00 00 00 00. Putting this hex number into the conversion tool (don't forget that you are on a little endian machine!) gives 5.308E-315. Printing this very small value with just 2 digits after the decimal point is 0.00.

HTH, Andreas