Answer SL3

main   : a=1     b=2
block1 : a=1     b=3
block2s: a=1     b=2
block2e: a=4     b=2
main   : a=1     b=2

#include <stdio.h>
int main( void )
{
  int a = 1;
  int b = 2;
  
  printf("main   : a=%d\tb=%d\n", a, b );
  
  {
    int b = 3;
    printf("block1 : a=%d\tb=%d\n", a, b );
  }
  
  {
    printf("block2s: a=%d\tb=%d\n", a, b );

    int a = 4;
    printf("block2e: a=%d\tb=%d\n", a, b );
  }
  
  printf("main   : a=%d\tb=%d\n", a, b );
  
  return 0;
}

Comments: Each inner block declares a variable that overshadows a variable declared by the main block. However, the scope of those inner variables is limited to the block in which they were declared.

In the second nested block, the variable a in the first printf statement refers to the variable in the outer block. Then the block declares another variable a which has a scope starting at that point. The second printf statement in that block refers to that new variable.

Question: How many identifiers are in the above program? There are two identifiers: a and b.

Question: How many variables are in the above program? There are four variables: two int variables named a and b in the outer scope, an int variable named b in the first nested scope, and an int variable named a in the second nested scope.

When a change is made to a variable, no other variable is affected, not even variables with the same name in other scopes.