Q: I have a function
extern int f(int *);which accepts a pointer to an int. How can I pass a constant by reference? A call like
f(&5);doesn't seem to work.
A: In C99, you can use a ``compound literal'':
Prior to C99, you couldn't do this directly; you had to declare a temporary variable, and then pass its address to the function:
int five = 5; f(&five);In C, a function that accepts a pointer to a value (rather than simply accepting the value itself) probably intends to modify the pointed-to value, so it may be a bad idea to pass pointers to constants. [footnote] Indeed, if f is in fact declared as accepting an int *, a diagnostic is required if you attempt to pass it a pointer to a const int. (f could be declared as accepting a const int * if it promises not to modify the pointed-to value.)
See also questions 2.10, 4.8, and 20.1.