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

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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?
implicit-conversion c
2 Answers

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).

user_23808300

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.

user_39238960
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