comparing int with size_t

Question

If i have a int and a size_t variable,can i compare them like:

int i=1;
size_t y=2;
if(i==y)
do something..

or i have to type-cast one of them?

Answer 1

size_t is going to be some sort of integer type (although possibly unsigned, so it might generate a warning) so the appropriate casting should be done for you automatically.

As others have already said, you may want to revisit whatever calculation is producing the int and see if you can do it in size_t in the first place if you're computing a required size for something.

Answer 2

It is okay to compare a size_t value with an int value, the int value will be implicitly converted to unsigned type.

Some compilers will issue a warning when you mix signed and unsigned types in comparisons. In that case you can explicitly convert the signed value to an appropriate unsigned type to suppress the warning.

Answer 3

It's safe provided the int is zero or positive. If it's negative, and size_t is of equal or higher rank than int, then the int will be converted to size_t and so its negative value will instead become a positive value. This new positive value is then compared to the size_t value, which may (in a staggeringly unlikely coincidence) give a false positive. To be truly safe (and perhaps overcautious) check that the int is nonnegative first:

/* given int i; size_t s; */
if (i>=0 && i == s)

and to suppress compiler warnings:

if (i>=0 && (size_t)i == s)