Why does the .second reference of const std::pair& return a copy of second value rather than a const reference?

I’m trying to get const pointers to each value in std::map. I tried const pair<...>& with its .second attribute but got wrong pointers. However, the const_ierator goes right.

A piece of simple toy code:

const map<int, double> intMap = {{1, 3.3}, {4, 5.7}, {2, 8.314}}; 
const double* cPtr; 
for (const pair<int, double>& cPair: intMap)
{
    cPtr = &(cPair.second); 
    cout << cPtr << endl; 
}
    
cout << endl; 
for (
    map<int, double>::const_iterator mapIter = intMap.begin(); 
    mapIter != intMap.end(); 
    mapIter ++
)
    cout << &(mapIter -> second) << endl; 

Which returns:

0x7ffc553c8210 # wrong pointers
0x7ffc553c8210
0x7ffc553c8210

0xb3ced8 # which I want
0xb3cf58
0xb3cf18

My questions:

• Why can’t I (or How can I) force the .second of const pair<>& to return a reference rather than a copy?
• If I cannot get what I want with const pair<>&, is there any more convenient way to get value references rather than using const_iterator, which seems really stupid?

>Solution :

Why can’t I (or How can I) force the .second of const pair<>& to return a reference rather than a copy?

Copy doesn’t happen with .second, but with mismatching type for std::pair.

Deferencing a map (const) iterator returns a /*const*/std::pair<const Key, Value>& (note the extra const).

When you do

for (const pair<int, double>& cPair: intMap)

You have a temporary pair<int, double> created from pair<const int, double> (that is why you see the same address).

If I cannot get what I want with const pair<>&, is there any more convenient way to get value references rather than using const_iterator, which seems really stupid?

You can with proper type:

for (const std::pair<const int, double>& cPair : intMap)

or with auto being less explicit but avoid any typo:

for (const auto& cPair : intMap)

or in C++17, with struture binding

for (const auto& [i, d] : intMap)