How to create a typedef for function pointers

c++function-pointerstypedef

I think it would be easier to use function pointers if I created a typedef for a function pointer, but I seem to be getting myself tripped up on some syntax or usage or something about typedef for function pointers, and I could use some help.

I've got

int foo(int i){ return i + 1;}
typedef <???> g;
int hvar;
hvar = g(3)

That's basically what I'm trying to accomplish I'm a rather new C programmer and this is throwing me too much. What replaces <???> ?

Best Solution

Your question isn't clear, but I think you might want something like this:

int foo(int i){ return i + 1;}

typedef int (*g)(int);  // Declare typedef

g func = &foo;          // Define function-pointer variable, and initialise

int hvar = func(3);     // Call function through pointer

Setting a bit

Use the bitwise OR operator (|) to set a bit.

number |= 1UL << n;

That will set the nth bit of number. n should be zero, if you want to set the 1st bit and so on upto n-1, if you want to set the nth bit.

Use 1ULL if number is wider than unsigned long; promotion of 1UL << n doesn't happen until after evaluating 1UL << n where it's undefined behaviour to shift by more than the width of a long. The same applies to all the rest of the examples.

Clearing a bit

Use the bitwise AND operator (&) to clear a bit.

number &= ~(1UL << n);

That will clear the nth bit of number. You must invert the bit string with the bitwise NOT operator (~), then AND it.

Toggling a bit

The XOR operator (^) can be used to toggle a bit.

number ^= 1UL << n;

That will toggle the nth bit of number.

Checking a bit

You didn't ask for this, but I might as well add it.

To check a bit, shift the number n to the right, then bitwise AND it:

bit = (number >> n) & 1U;

That will put the value of the nth bit of number into the variable bit.

Changing the nth bit to x

Setting the nth bit to either 1 or 0 can be achieved with the following on a 2's complement C++ implementation:

number ^= (-x ^ number) & (1UL << n);

Bit n will be set if x is 1, and cleared if x is 0. If x has some other value, you get garbage. x = !!x will booleanize it to 0 or 1.

To make this independent of 2's complement negation behaviour (where -1 has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.

number ^= (-(unsigned long)x ^ number) & (1UL << n);

unsigned long newbit = !!x;    // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);

It's generally a good idea to use unsigned types for portable bit manipulation.

number = (number & ~(1UL << n)) | (x << n);

(number & ~(1UL << n)) will clear the nth bit and (x << n) will set the nth bit to x.

It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.

C++ – Difference between ‘struct’ and ‘typedef struct’ in C++

In C++, there is only a subtle difference. It's a holdover from C, in which it makes a difference.

The C language standard (C89 §3.1.2.3, C99 §6.2.3, and C11 §6.2.3) mandates separate namespaces for different categories of identifiers, including tag identifiers (for struct/union/enum) and ordinary identifiers (for typedef and other identifiers).

If you just said:

struct Foo { ... };
Foo x;

you would get a compiler error, because Foo is only defined in the tag namespace.

You'd have to declare it as:

struct Foo x;

Any time you want to refer to a Foo, you'd always have to call it a struct Foo. This gets annoying fast, so you can add a typedef:

struct Foo { ... };
typedef struct Foo Foo;

Now struct Foo (in the tag namespace) and just plain Foo (in the ordinary identifier namespace) both refer to the same thing, and you can freely declare objects of type Foo without the struct keyword.

The construct:

typedef struct Foo { ... } Foo;

is just an abbreviation for the declaration and typedef.

Finally,

typedef struct { ... } Foo;

declares an anonymous structure and creates a typedef for it. Thus, with this construct, it doesn't have a name in the tag namespace, only a name in the typedef namespace. This means it also cannot be forward-declared. If you want to make a forward declaration, you have to give it a name in the tag namespace.

In C++, all struct/union/enum/class declarations act like they are implicitly typedef'ed, as long as the name is not hidden by another declaration with the same name. See Michael Burr's answer for the full details.