C++ – How to get a method’s address in a class

Here is part of my code:

In a.h:

class classA
{
public:
void (*function_a)(void);
classA();
void function_a();
};

In a.cpp:

void classA::classA()
{
  (*function_a)() = function_a;
}

void classA::function_a()
{
  return;
}

I want to get function_a's address and save it into void (*function_a)(void), but I got compile error that "expression is not assignable". What shall I do to solve this problem?

Best Answer

First of all, choose different names for different things.

Second, the non-static member function pointer should be declared as:

void (classA::*memfun)(void); //note the syntax

then the assignment should be as:

memfun = &classA::function_a; //note &, and note the syntax on both sides.

and you call this as:

classA instance;
(instance.*memfun)();

That is what you do in C++03.

However, in C++11, you can use std::function as well. Here is how you do it:

std::function<void(classA*)> memfun(&classA::function_a);

and you call this as:

classA instance;
memfun(&instance);

Online demo

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++ – How to iterate over the words of a string

I use this to split string by a delimiter. The first puts the results in a pre-constructed vector, the second returns a new vector.

#include <string>
#include <sstream>
#include <vector>
#include <iterator>

template <typename Out>
void split(const std::string &s, char delim, Out result) {
    std::istringstream iss(s);
    std::string item;
    while (std::getline(iss, item, delim)) {
        *result++ = item;
    }
}

std::vector<std::string> split(const std::string &s, char delim) {
    std::vector<std::string> elems;
    split(s, delim, std::back_inserter(elems));
    return elems;
}

Note that this solution does not skip empty tokens, so the following will find 4 items, one of which is empty:

std::vector<std::string> x = split("one:two::three", ':');