Question

Code:

int main()  
{
  int a=1;
  switch(a)
  {
    int b=20;

    case 1:
    printf("b is %d\n",b);
    break;

    default:
    printf("b is %d\n",b);
    break;
  }
  return 0;
}

Output:
It prints some garbage value for b
when does the declaration of b takes place here
Why b is not initialized with 20 here???

Answer 1

Because memory will be allocated for int b but when the application is run “b = 20” will never be evaluated.

This is because your switch-statement will jump down to either case 1:¹ or default:, skipping the statement in question – thus b will be uninitialized and undefined behavior is invoked.

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The following two questions (with their accepted answers) will be of even further aid in your quest searching for answers:

• How can a variable be used when it’s definition is bypassed? ²

• Why can’t variables be declared in a switch statement?

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Turning your compiler warnings/errors to a higher level will hopefully provide you with this information when trying to compile your source.

Below is what gcc says about the matter;

foo.cpp:6:10: error: jump to case label [-fpermissive]
foo.cpp:5:9: error:   crosses initialization of 'int b'

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¹ since int a will always be 1 (one) it will always jump here.

² most relevant out of the two links, answered by me.