doubt on function pointer using typedef

**krkr** · 11-03-2015

As per my understanding following is the syntax for function pointer using typedef.

Code:

typedef void (*fp)(int arg);

I saw typedef to function definition similar to below.

Code:

typedef int func(int arg);
func *fa[] = {
    [0] = f1,
    [1] = f2,
    [2] = f3,
    [3] = f4,
    [4] = f5,
};

Is it array of function pointers and the typedef declaration in the above code snippet is a function pointer or different.

I got the dobht after seeing the seeing the declarations @
git.denx.de Git - u-boot.git/blob - common/cmd_bootm.c

git.denx.de Git - u-boot.git/blob - common/cmd_bootm.c

**Codeplug** · 11-03-2015

It's basically a function declaration without the pointer part. Notice the star in "func *fa[] = ...". So it an array of pointers to func.
It's also used to just declare functions using "variable style syntax":

Code:

static boot_os_fn do_bootm_vxworks;
static boot_os_fn do_bootm_qnxelf;

gg

**krkr** · 11-03-2015

I couldn't understand

"So it an array of pointers to func.
It's also used to just declare functions using "variable style syntax":"

Can you please elaborate a bit.

**Codeplug** · 11-03-2015

typedef void (*fp)(int arg);
This is typedef of a pointer to a function fp.

typedef void f(int arg);
This is a typedef of function f (no pointer).

Code:

f foo; // declaration of a function foo
fp pfoo; // definition of a function pointer pfoo
f *pfoo2; // definition of function pointer pfoo2

void foo(int arg) {} // definition of function foo

pfoo = pfoo2 = &foo; // both point to foo()

gg

**Hodor** · 11-06-2015

Personally, if it was my program, I just wouldn't typedef the pointer to a function (IMO it makes things confusing and hard to read)

**whiteflags** · 11-06-2015

I find not typedefing function pointers to be silly - they aren't used like normal pointers, otherwise you might have a point there. No matter what function-like thing foo is, foo() is an invocation.

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