C++: How to call by pointer without ampersand? [closed] - c++

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How would one go about calling a function by pointer without using the ampersand in the function call where I pass in a struct. I can only edit the function definition, not the way the function is being called. Call by reference can be used, but I would like to call by pointer, is there any way of doing this?
void Some_Func(struct TheStruct &My_struct){
My_struct->member = 4; // Would like to do this!
}
int main(){
Some_Func(My_Struct);
}

Answer 1

To be honest, I think this question was triggered by some misunderstanding....Lets assume you have this (working) code:
struct TheStruct { int member;};
void Some_Func(struct TheStruct &My_struct){
// some implementation
}
int main(){
TheStruct x;
Some_Func(x);
}
You say
I can only edit the function definition, not the way the function is being called.
...thus there is no way to make the function take a pointer as parameter. If there is a instance (not a pointer) passed as parameter in main and you are not allowed to change main, then you cannot magically make the parameter a pointer.
The only way to use -> inside the function is
void Some_Func(struct TheStruct &My_struct){
TheStruct* x = &MyStruct;
x->member = 4;
}
However, there is not a single reason to do so (unless the . on your keyboard is broken) and it makes the code confusing to read.

Answer 2

I am 90% sure this is not the answer you are looking for, but I'll post it for the remaining 10% of doubt anyway:
Have a look at std::addressof(T&). It gives you the pointer without using an ampersand.

Answer 3

When accessing an object by reference, you have to use the . operator to reach its members:
void Some_Func(TheStruct &My_struct) {
My_struct.member = 4;
}
int main() {
TheStruct My_struct;
Some_Func(My_Struct);
// or:
// TheStruct *My_struct = new My_struct;
// Some_Func(*My_Struct);
// delete My_struct;
}
When accessing an object by pointer, you have to use the -> or *. operator to reach its members:
void Some_Func(TheStruct *My_struct) {
My_struct->member = 4;
// or (*My_struct).member = 4;
}
int main() {
TheStruct My_struct;
Some_Func(&My_Struct);
// or:
// TheStruct *My_struct = new TheStruct;
// Some_Func(My_Struct);
// delete My_struct;
}
If you don't want to use the & operator to get the address of an object variable (maybe because it overrides operator&, for instance), you can use std::addressof() in C++11 and later:
int main() {
TheStruct My_struct;
Some_Func(std::addressof(My_Struct));
}
Or use boost::addressof:
#include <boost/utility.hpp>
int main() {
TheStruct My_struct;
Some_Func(boost::addressof(My_Struct));
}
Or implement addressof() manually:
template<class T>
T* my_addressof(T& arg)
{
return reinterpret_cast<T*>(
&const_cast<char&>(
reinterpret_cast<const volatile char&>(arg)
)
);
}
int main() {
TheStruct My_struct;
Some_Func(my_addressof(My_Struct));
}

Answer 4

This will work.
struct TheStruct {
int member;
};
void Some_Func(TheStruct &My_struct){
My_struct.member = 4;
}
int main(){
TheStruct My_Struct;
My_Struct.member = 5;
std::cout << My_Struct.member << std::endl;
Some_Func(My_Struct);
std::cout << My_Struct.member << std::endl;
return 0;
}
Output:
5
4