C++ Confusing scope of redeclared variables in for loop

1
votes

The following while loop does not terminate. This is because the variable x is being re-declared inside the while loop. But I don't understand why in the second iteration onward, the statements x<10 and y=x considers the x defined in the outer scope and not the x defined in the block scope in the following statement. Is this because once the first iteration ends, the x defined in the block scope is destroyed and the loop begins to execute fresh? 

#include<iostream>
int main () {
  int x = 0, y;  
  while(x <10 ){
    y = x;
    std::cout<<"y is :"<< y <<std::endl;  
    int x = y + 1;
    std::cout<<"x is :"<< x <<std::endl;  
  }
  std::cout<<"While loop is over"<<std::endl;
}
2
c++while-loopscoperedeclaration

2 Answers

4
votes

Every iteration the while loop evaluates the outer scope x and y is assigned the value of the outer scope x. After that another x is defined in the inner scope which is what the second std::cout uses, but the program makes no other use of the inner x

In the code below I replaced the inner x with z but otherwise the behavior is identical. The only difference is that there is not a second x within a more inner scope to hide the outer one:

#include<iostream>

int main () {
    int x = 0, y;
    while(x <10 ){
        y = x;
        std::cout<<"y is :"<< y <<std::endl;
        int z = y + 1;
        std::cout<<"z is :"<< z <<std::endl;
    }
    std::cout<<"While loop is over"<<std::endl;
}

Below I have an example that is intended to clear the confusion. In the inner scope x is not being "re-declared", a new x is being declared and it goes out of scope after the }:

#include<iostream>

int main () {
    int x = 1;
    {
        int x = 2;
        std::cout << x << '\n'; // 2
    }
    std::cout << x << '\n'; // 1
}
2
votes

Yes, you understand it correctly. So every time when compare in while, it is using the outer x.

while (x < 10) {
    y = x; //Here the x is the outer one. The inner one does not exist yet.
    std::cout << "y is :" << y << std::endl;  
    int x = y + 1; // From here, x will refer to the inner one.
    std::cout << "x is :" << x << std::endl;  
    // inner x is destroyed.
}