https://gcc.gnu.org/bugzilla/show_bug.cgi?id=112101
Bug ID: 112101
Summary: feature request: typeof_arg for extracting the type of
a function's (or function pointer's) arguments
Product: gcc
Version: unknown
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c
Assignee: unassigned at gcc dot gnu.org
Reporter: malekwryyy at gmail dot com
Target Milestone: ---
C23 will add typeof (although gcc had a for a while) which gives the type of an
expression or a type. By using it, it is possible to get the return type of a
function like so:
```
int func();
typeof(func()) x; // int x;
```
But there's no way to extract the type of the argument of a function:
```
void func(int);
?? x;
```
I think something like 'typeof_arg' would be a good addition.
It takes 2 operands, first is function or function pointer, and second is an
integer constant for the index of the argument, which must be within [0,
arg_count).
for example:
```
#define print_func(f) \
printf(#f \
"(" \
_Generic( (__typeof_arg(f, 0)){0}, \
int: "int", \
long: "long", \
float: "float", \
char*: "char*", \
default: "other ") \
")")
```
this would print a single-argument function's name and arg type like this
"puts(char*)".
another example:
```
#define gurantee_type(exp, type) \
_Generic(exp, type: exp, default: (typeof(exp)){0})
#define call_with_empty(f) \
_Generic( (__typeof_arg(f, 0)){0}, \
char*: gurantee_type(f, void(*)(char*))(""), \
default: f( (__typeof_arg(f, 0)){0} ) \
)
```
which calls the function 'f' with empty string if it takes char*, or 0 of the
correct type otherwise.
this wouldn't work for variadic functions, so __typeof_arg(printf, 1) would be
an error.
I think a feature like this would be really helpful for generic programming in
C
