What is the need for a C_INT fortran type? Are C and Fortran integers so different?

Question

If I declare an integer in fortran as: INTEGER(C_INT) :: i then, if I understand correctly, it's safe to be passed to a C function. Now disregarding the added headache of always declaring integers this way, is there any reasons not to always declare your variables as C-interoperable? Is there any downside to doing so?

Also, in the case of something as simple a an integer, what exactly does the C_INT change from a traditional Fortran integer? Are Fortran and C integers actually different?

Answer 1

The C integer size is usually fixed because the OS is compiled using C and the system bindings are published as C headers. The operating system designers choose one of the common models like LLP64, LP64, ILP64 (for 64-bit pointer OS's) and the C compilers then follow this choice.

But Fortran compilers are more free. You can set them to a different configuration. Fortran compiler set to use 8-byte default integers and 8-byte default reals are still perfectly standard conforming! (C would be as well but the choice is fixed in the operating system.)

And because the integers in C and in Fortran do not have to match you need a mechanism to portably select the C interoperable kind, whatever the default kind is.

This is not just academic. you can find libraries like MKL compiled with 8-byte integers (the ILP64 model, which is not used by common operating systems for C).

So when you call some API, some function whose interface is defined in C, you want to call it properly, not depending on the settings of the Fortran compiler. That is the use case for the C-interoperable types. If a C function requires an int, you give it an integer(c_int) and you do not care if the Fortran default integer is the same or not.