output of the following program

Question

2.67/5 (3 votes)

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What will be the output of the program?

C

#include<stdio.h>
int main()
{
 float a = 0.7;
 if(0.7 > a)
 printf("Hi\n");
 else
 printf("Hello\n");
 return 0;
}

the answer is given as hi...and i was expecting it to be hello.please help.

Posted 10-Feb-15 23:49pm

Member 11324568

Updated 11-Feb-15 12:26pm

Andreas Gieriet

v2

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3 solutions

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Abhinav S · Accepted Answer · 2015-02-10T23:59:00

Go through this article - Comparing floating point numbers[^].

"Floating point math is not exact. Simple values like 0.2 cannot be precisely represented using binary floating point numbers, and the limited precision of floating point numbers means that slight changes in the order of operations can change the result. Different compilers and CPU architectures store temporary results at different precisions, so results will differ depending on the details of your environment. If you do a calculation and then compare the results against some expected value it is highly unlikely that you will get exactly the result you intended."

Thus you can try if (fabs(0.7 - a) < 0.00001)

CPallini · Accepted Answer · 2015-02-11T10:53:00

Solution 2

There is a subtlety in you program: it

Assigns a double constant (0.7 is a double literal) to a float variable (in this step rounding occurs).
Compares such a float variable with (the same) double constant.

Try the following code:

C

#include<stdio.h>
int main()
{
  float a = 0.7f;
  if(0.7f > a)
    printf("Hi\n");
  else
    printf("Hello\n");
  return 0;
}

as well as

C

#include<stdio.h>
int main()
{
  double a = 0.7;
  if (0.7 > a)
    printf("Hi\n");
  else
    printf("Hello\n");
  return 0;
}

both of them output "Hello" on my system.

Posted 11-Feb-15 10:53am

CPallini

Comments

Andreas Gieriet 11-Feb-15 19:51pm

~~On my system (VS2010) I get "Hi" only - i.e. not the "intuitive" solution...~~
Cheers
Andi

CPallini 12-Feb-15 2:57am

Both of the above programs produce "Hello", compiled with Visual Studio 2013 Express.

Andreas Gieriet 12-Feb-15 4:35am

Strange...
Any special rounding settings?
What is sizeof(double) and sizeof(float)? Mine is 8 and 4 respectively.
Cheers
Andi

CPallini 12-Feb-15 4:56am

Here the same: compilers cannot cheat on that.

Andreas Gieriet 12-Feb-15 5:26am

What do you get if you run my hack from solution 3?
Andi

CPallini 12-Feb-15 5:35am

The same output you got.

Andreas Gieriet 12-Feb-15 5:43am

Ah, my mistake! Sorry!
Of course, your code shows how to do it correctly to get the comparison right.
Sorry again for the confusion.
Andi

CPallini 12-Feb-15 7:00am

Oh, never mind.
Hope the OP got it all.

Member 11324568 12-Feb-15 3:42am

i guess thats compiler dependent ..thanks cpallini :)

CPallini 12-Feb-15 4:24am

I guess not.
By the way, you are welcome.

Andreas Gieriet · Accepted Answer · 2015-02-11T14:12:00

You may try the following hack (on a little endian system):

C

// allow to extract the bit pattern of the floating point value (little endian)
typedef union {
  double d;
  struct { unsigned long long m:52, e:11, s:1; } x;
} double_union;
// allow to extract the bit pattern of the floating point value (little endian)
typedef union {
  float f;
  struct { unsigned long m:23, e:8, s:1; } x;
} float_union;
// aux function to write the floating values
void write_bits(long e, unsigned long long m, int is_float)
{
  size_t b = is_float ? 23 : 52;
  long bias = is_float ? 127 : 1023;
  size_t n = sizeof(m)*8;
  unsigned long long mask = (0x1ull << (n-1));
  printf("%s.", !e && !m ? "0" : "1");
  m <<= (n-b);
  while(b-- > 0)
  {
    printf("%s", (m & mask) ? "1" : "0");
    m <<= 1;
  }
  if (e) printf(" 2^%d", e-bias);
}
// write the float as bit-mantissa
void write_float(float_union u)
{
  printf("%s", u.x.s ? "-" : "+");
  unsigned long e = u.x.e;
  unsigned long m = u.x.m;
  write_bits(e, m, 1);
  printf("\n");
}
// write the double as bit-mantissa
void write_double(double_union u)
{
  printf("%s", u.x.s ? "-" : "+");
  unsigned long e = u.x.e;
  unsigned long long m = u.x.m;
  write_bits(e, m, 0);
  printf("\n");
}
// test program
int main()
{
  float a = 0.7;
  if(0.7 > a) printf("Hi\n");
  else        printf("Hello\n");

  double_union du;
  float_union  fu;

  fu.f = 0.7;
  write_float(fu);
  du.d = 0.7;
  write_double(du);
  du.d = fu.f;  // <----- THIS IS THE TROUBLESOME PART: float assigned to double
  write_double(du);

  return 0;
}

The output on my little endian system with 4-byte float and 8-byte double is

Hi
+1.01100110011001100110011 2^-1
+1.0110011001100110011001100110011001100110011001100110 2^-1
+1.0110011001100110011001100000000000000000000000000000 2^-1

This shows that the two values are not equal anymore.
0.7 is of type double.
0.7 > a promotes the float a value to a double value. This results in the above mentioned different values. The double 0.7 is indeed greater than the float 0.7f since the float value is promoted to the double value for comparison.

What to learn from this: never mix number domains, especially for floating point numbers.

See also What Every Computer Scientist Should Know About Floating-Point Arithmetic[^].

Cheers
Andi

output of the following program

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