why can I still give const value a new value - c++

Question

The book said I cannot change the value of const once I gave it a number， but it seems I can still give it a number even if it was given.
#include<iostream>
using namespace std;
const int fansc(100);
cout<< fansc << endl; //output:100
int fansc(20);
cout<< fansc << endl;//output:20

Answer 1

The C++ code you gave won't compile, and rightly so. A const variable(a) is, well, ... constant. The error is shown in the following program and transcript:
#include <iostream>
using namespace std;
int main() {
const int fansc(100);
cout << fansc << endl;
int fansc(20);
cout << fansc << endl;
}
pax> g++ --std=c++17 -Wall -Wextra -Wpedantic -o prog prog.cpp
prog.cpp: In function ‘int main()’:
prog.cpp:6:9: error: conflicting declaration ‘int fansc’
6 | int fansc(20);
| ^~~~~
prog.cpp:4:15: note: previous declaration as ‘const int fansc’
4 | const int fansc(100);
| ^~~~~
That leaves the Anaconda bit that you mention in a comment. I have little experience with that but it seems to me the only way that would work is if the second fansc definition was somehow created in a different scope to the first. In real C++ code, that would go something like:
#include <iostream>
using namespace std;
int main() {
const int fansc(100);
cout << fansc << endl;
{ // new scope here
int fansc(20);
cout << fansc << endl;
} // and ends here
cout << fansc << endl;
}
And the output of that is:
pax> g++ --std=c++17 -Wall -Wextra -Wpedantic -o prog prog.cpp && ./prog
100
20
100
(a) Yes, I know that's a self-contradiction :-)