Strange behaviour of variable in c++

Question

I am unable to understand how a variable defined in an outer scope can be redefined in an inner scope.

Here is my sample code. The code compiles successfully.

#include<iostream>
#include<map>
using namespace std ;

#define ll long long int

int main(int argc, char const *argv[])
{
    ll t;
    cin>>t;
    while(t--)
    {
        map<ll,ll> mp;
        map<ll,ll>::iterator t=mp.end();
    }
    return 0;
}

What are you confused about exactly? It just can. Them's the rules. What do you want to know? — Asteroids With Wings, Apr 30 '20 at 19:40
When there are two variables with the same name available within a scope, the compiler will use the one declared in the most-inner scope. (If you turn on warnings, it'll probably give you a warning about shadowing, though, since accidentally declaring and using an inner-scope variable when you intended to use an outer-scope variable with the same name is a somewhat common programmer mistake) — Jeremy Friesner, Apr 30 '20 at 19:42
"Shadowing" a variable is perfectly legal, if albeit *usually* undesirable. — Remy Lebeau, Apr 30 '20 at 19:47
its not "redefining" you declare a different variable that happens to have the same name — 463035818_is_not_an_ai, Apr 30 '20 at 20:08

score 0 · Accepted Answer · answered Apr 30 '20 at 23:22

0

It's not the same variables. C++ use the variable that defined in the scope first before globals variable in the same name. So in this example, you have two variables in the loop: long long t global and t map.

answered Apr 30 '20 at 23:22

jacob galam

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Strange behaviour of variable in c++

1 Answers1