Why didn't the value of x change when the condition is true?

Question

int x = 1;
if (x > 0)
    int x = 2;
cout << x;

I expected the output would to be 2 because the condition is true, but what happens here?

I received 1 as output.

Answer 1

You've shadowed the variable. This occurs when a variable declared within one scope has the same name as a variable declared in an outer scope.

Making this modification will give the output you expect:

int x = 1;
if (x > 0) {
  x = 2; // now you're modifying the same x
}
cout << x;

Answer 2

In c++ variable or other symbol declarations are local to their scope, redeclaring them in a different scope is called shadowing, and you will see no effect of the assignment outside of your current scope level. This concerns

• Local scope (everything within braces {} or immediately appearing after a control flow statement like if, else, case, while or for)
  Typical examples:

  int i = 5;
  if(i == 5)
      int i = 2; // Single statement scope following the if
                 // Changes the local variable's value
  

  ---

  int j = 42;
  if(j == 42) {
      int j = 2; // Scoped block local variable
                 // Changes the local variable's value
  }
  

• Class scope (any class member variables)
  Typical examples:

  class MyClass {
      int myMember_;
  public:
      MyClass(int aValue) {
          int myMember_; // another local variable in the constructor function
          myMember_ = aValue; // Changes the local variable's value
      }
  };
  

  ---

  class MyClass {
      int myMember_;
  public:
      MyClass(int aValue) {
          int myMember_ = aValue; // Changes the local variable's value
      }
  };
  

• Namespace scope (any namespace global variables).
  Same principle as above. Namespaces can shadow variables (symbols) from other namespaces

---

Your c++ compiler should probably give you a warning about the appearance of one of the above mentioned situations.
With specifically GCC (g++) you can force that using the -Wshadow compiler flag.