C++ Member Functions (getters) - c++

Question

I've seen people define a member function like this:
void getValue(int& v)
{
v = m_value;
}
and also like this:
int getValue()
{
return m_value;
}
I guess the first saves memory? Is that the only time you would use the first type of get-function? The second seems a lot more convenient.

Answer 1

I thought I would godbolt it for you
source
#include <iostream>
struct Foof{
int m_val;
Foof(int v){
m_val = v;
}
void woodle()
{
if(m_val > 42)
m_val++;
else
m_val--;
}
void Get1(int &v)
{
v = m_val;
}
int Get2()
{
return m_val;
}
};
int main(int c, char**v){
int q;
std::cin >> q;
Foof f1(q);
std::cin >> q;
Foof f2(q);
f1.woodle();
f2.woodle();
int k;
f1.Get1(k);
int j = f2.Get2();
std::cout << k << j;
}
the woodle function and the cin to initialize is to make the compiler think a bit
I have 2 foofs otherwise the compiler goes "well I know the answer to this question" when I call Get2 after Get1
compiled with -03 - ie optimize hard. The code comes out as (gcc)
pushq %rbx
movl $_ZSt3cin, %edi
subq $16, %rsp
leaq 12(%rsp), %rsi
call std::basic_istream<char, std::char_traits<char> >::operator>>(int&)
movl 12(%rsp), %ebx
leaq 12(%rsp), %rsi
movl $_ZSt3cin, %edi
call std::basic_istream<char, std::char_traits<char> >::operator>>(int&)
movl 12(%rsp), %eax
movl $_ZSt4cout, %edi
leal 1(%rbx), %edx
cmpl $43, %ebx
leal -1(%rbx), %esi
cmovge %edx, %esi
leal -1(%rax), %ebx
leal 1(%rax), %edx
cmpl $43, %eax
cmovge %edx, %ebx
call std::basic_ostream<char, std::char_traits<char> >::operator<<(int)
movq %rax, %rdi
movl %ebx, %esi
call std::basic_ostream<char, std::char_traits<char> >::operator<<(int)
addq $16, %rsp
xorl %eax, %eax
popq %rbx
ret
I separated out the actual calls to Get1 or Get2 you can see that
the generated code is identical
the compiler is very aggressive at optimizing, there are no function calls etc
Lesson, write your code to be human readable and let the compiler do the heavy lifting