Re: Better way to call member function using function pointer?
Ganesh Pagade wrote:
Following is isolated code reproducing the issue I encountered while
using function pointers:
//////////////////////////////////////////////
#include <iostream>
#include <string>
#include <map>
using namespace std;
class Sample
{
public:
Sample()
{
funPtrs["fun1"] = &Sample::fun1;
funPtrs["fun2"] = &Sample::fun2;
funPtrs["fun3"] = &Sample::fun3;
}
void process()
{
string str[] = { "fun1", "fun2", "fun3" };
// Method 1 - Compilation error.
for (int i = 0; i < 3; ++i)
{
// VS 2005 gives me error as:
// error C2064: term does not evaluate to a function taking 1
arguments
funPtrs[str[i]]("hi");
}
// Method 2 - Works fine.
for (int i = 0; i < 3; ++i)
{
ptr = funPtrs[str[i]];
(this->*ptr)("hello");
}
}
private:
void fun1(string str)
{
str = "fun1";
cout << str << endl;
}
void fun2(string str)
{
str = "fun2";
cout << str << endl;
}
void fun3(string str)
{
str = "fun3";
cout << str << endl;
}
private:
static map<string, void (Sample::*)(string)> funPtrs;
void (Sample::*ptr)(string);
};
map<string, void (Sample::*)(string)> Sample::funPtrs;
int main()
{
Sample s;
s.process();
return 0;
}
//////////////////////////////////////////////
To my understanding Method 1 gives error because function is called
using pointer which is not a member of Sample, due to which "this"
cannot be passed to the member function as argument. Hence the
mentioned error.
No. This has absolutely nothing to do with the pointer 'ptr' being a
member or not, i.e. there's absolutely no need to introduce a member
'ptr' into your class.
The first "method" does not work simply because in C++ pointers of
'pointer-to-member' type cannot be dereferenced by '->' and '*'
operators. Instead, they have to be dereferenced with '->*' and '.*'
operators. Note, that '->*' and '.*' are completely independent
operators, different from '->' and '*'.
Pointers to members cannot be dereferenced "by themselves", since
conceptually they are not really pointing into any concrete locations.
Think of them as of "relative" pointers: they are relative to a concrete
object. (For example, in case of pointers to _data_ members, you can
think of them as _offsets_ from the beginning of the object to the
actual data member.)
So, in order to dereference a pointer of 'pointer-to-member' type you
have to supply an additional operand: a concrete object, relative to
which this pointer-to-member is supposed to be used. The object is
specified on the left-hand side of '->*'/'.*' operators.
This is what you did in your second method: you used '->*' operator with
'this' on the left-hand side. So, the pointer 'ptr' is used in
conjunction with '*this' object. Note, again, that this has nothing to
do with 'ptr' being a member of class 'Sample'. You can easily make
'ptr' a local variable
for (int i = 0; i < 3; ++i)
{
void (Sample::*ptr)(string) = funPtrs[str[i]];
(this->*ptr)("hello");
}
or you can do it without any additional variables at all
for (int i = 0; i < 3; ++i)
(this->*funPtrs[str[i]])("hello");
(This is how your "Method 1" should have looked from the very beginning).
--
Best regards,
Andrey Tarasevich
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