#include <stdio.h> int main(void){ float a = 1.1; double b = 1.1; if(a == b){ printf("if block"); } else{ printf("else block"); } return 0; } Prints: else block
#include <stdio.h> int main(void){ float a = 1.5; double b = 1.5; if(a == b){ printf("if block"); } else{ printf("else block"); } return 0; } Prints: if block
What is the logic behind this?
Compiler used: gcc-4.3.4
This is because 1.1 is not exactly representable in binary floating-point. But 1.5 is.
As a result, the float and double representations will hold slightly different values of 1.1.
Here is exactly the difference when written out as binary floating-point:
(float) 1.1 = (0.00011001100110011001101)₂ (double)1.1 = (0.0001100110011001100110011001100110011001100110011010)₂ Thus, when you compare them (and the float version gets promoted), they will not be equal.
The exact value of 1.1 decimal in binary is non-ending fraction 1.00011001100110011001100(1100).... The double constant 1.1 is 53-bit truncation / approximate value of that mantissa. Now this when converted to float, the mantissa will be represented just in 24 bits.
When the float is converted back to double, the mantissa is now back to 53 bits, but all memory of the digits beyond 24 are lost - the value is zero-extended, and now you're comparing (for example, depending on the rounding behaviour)
1.0001100110011001100110011001100110011001100110011001 and
1.0001100110011001100110000000000000000000000000000000 Now, if you used 1.5 instead of 1.1;
1.5 decimal is exactly 1.1 in binary. It can be presented exactly in just 2 bit mantissa, therefore even the 24 bits of float are an exaggeration... what you have is
1.1000000000000000000000000000000000000000000000000000 and
1.10000000000000000000000 The latter, zero extended to a double would be
1.1000000000000000000000000000000000000000000000000000 which clearly is the same number.
-Wfloat-equalin your compiler. It's a useful warning.