CODEWITHSUNDEEP
cimplicit-conversion
rpj
rpj

Reputation: 93

What happens when an int and float variable both having same value is compared using comparison operator?

What happens when an int and float variable both having same value is compared using comparison operator? 

 main( )
  {
    int x = 3 ;
    float y = 3.0 ;
    if ( x == y )
      printf ( "\nx and y are equal" ) ;
    else
      printf ( "\nx and y are not equal" ) ;
   }




output : x and y are equal
What happens when x is compared with y variable?

Upvotes: 3

Views: 2466

Answers (2)

Steve Summit
Steve Summit

Reputation: 48023

A good, general rule is that you should never compare floating-point numbers for exact equality. Famously, simple-looking code fragments like

float f1 = 1.1, f2 = 2.2;
float f3 = 3.3;
if(f1 + f2 == f3) printf("expected\n");
else printf("surprise\n");

are quite likely to behave surprisingly. The reason is that most real numbers (including ordinary decimal fractions like 1.1) can not be represented exactly in a finite-width floating-point representation.

Some numbers can be represented exactly, of course. In particular, small integers can typically be represented exactly even in low-precision floating-point representations. So to restate your example, code like

int x = 3;
float y = 3.0;
if(x == y) printf("expected\n");
else printf("surprise\n");

is virtually guaranteed to print the expected result on any practical computer.

How does it work, exactly? C supports mixed-mode arithmetic. You're perfectly allowed to write things like x == y even if x and y have different types. What typically happens is that the compiler inserts implicit conversions to a common type. So when you write x == y, the compiler says to itself. "Hmm. I don't have an instruction to compare an int and a float for equality. But if I convert the int to a float, I'll have two floats, and then I can use the machine's single-precision floating-point equality comparison operator. So I'll pretend the expression was (float)x == y."

But the general rule still stands. The fact that a floating-point variable can exactly store a small integer like 3.0 (or an exact fraction like 0.5) does not mean that it can store all integers exactly. For example, the similar-looking code

long int x = 123456789;
float y = 123456789.0;
if(x == (long int)y) printf("expected\n");
else printf("surprise\n");

is again quite likely to print a surprising result.

Upvotes: 2

Bathsheba
Bathsheba

Reputation: 234825

The int is converted implicitly to a float type; your code is equivalent to

if ((float)x == y)

Note this happens even if the conversion from an int to a float loses precision (which it doesn't in your case).

Upvotes: 5

Related Questions