Passing a constant integer when function expects a pointer

Question

What's the best/most cannonical way of passing in a constant integer value to a function that expects a pointer?

For example, the write function

write (int filedes, const void *buffer, size_t size);

Let's say I just want to write a single byte (a 1), I would think this:

write (fd, 1, 1);

but I obviously get the warning

warning: passing argument 2 of 'write' makes pointer from integer without a cast

---

I know I can do

int i = 1;
write (fd, &i, 1);

but is that necessary? What's the most correct way of doing this without the need of declaring/initializing a new variable?

Answer 1

In C89/90 you can't generally do it without declaring a new variable. The thing about pointers in C is that they always point to lvalues. Lvalue is something that has a location in memory. So, in order to supply the proper argument to your function, you need something that has a location in memory. That normally requires a definition, i.e. an object, a variable. Constants in C are not lvalues, so you can't use a constant there. Things that lie somewhat in between constants and variables are literals. C89/90 has only one kind of literal: string literal. String literals are lvalues, but not variables. So, you can use a string literal as the second argument of write.

In C99 the notion of literal was extended beyond string literals. In C99 you can also use compound literals for that purpose, which also happen to be lvalues. In your example you can call your function as

write(fd, (char[]) { 1 }, 1)

But this feature is only available in C99.