Trevor Hickey Trevor Hickey - 5 months ago 30
C Question

Why does printf("%f",0); give undefined behavior?

The statement

prints 0.

However, the statement
prints random values.

I realize I'm exhibiting some kind of undefined behaviour, but I can't figure out why specifically.

A floating point value in which all the bits are 0 is still a valid
with value of 0.

are the same size on my machine (if that is even relevant).

Why does using an integer literal instead of a floating point literal in
cause this behavior?


The "%f" format requires an argument of type double. You're giving it an argument of type int. That's why the behavior is undefined.

The standard does not guarantee that all-bits-zero is a valid representation of 0.0 (though it often is), or of any double value, or that int and double are the same size (remember it's double, not float), or, even if they are the same size, that they're passed as arguments to a variadic function in the same way.

It might happen to "work" on your system. That's the worst possible symptom of undefined behavior, because it makes it difficult to diagnose the error.

N1570 paragraph 9:

... If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.

Arguments of type float are promoted to double, which is why printf("%f\n",0.0f) works. Arguments of integer types narrower than int are promoted to int or to unsigned int. These promotion rules (specified by N1570 paragraph 6) do not help in the case of printf("%f\n", 0).