I read that initialization within cases of switch is supposed to give compiler error. But when I tried two different version. One is not giving out the error (Ex1), I can't understand why.
Ex 1:
switch (i)
{
case 1:
int k;
break;
case 2:
int j=3;
break;
}
Ex2:
switch (i)
{
case 1:
int k=3;
break;
case 2:
int j;
break;
}
In a switch block, every declaration that comes before a case is available at that case. The error occurs when there is a case to which the switch can jump to which is located after the initialization of a variable that you could access there. If that case was jumped to, you could use the variable despite its initialization never occuring.
In the first example, there is no case which would skip an initialization. The only initialization is int j=3; which comes after every case so it can't be skipped. k is not initialized so its initialization can't be skipped.
In the second example,if i == 2 then k's initialization would be skipped.
One solution is to define a scope for each case. Then, the variables defined for a case is only accessible from that case and the problem can't occur. For example :
void foo(int i)
{
switch (i)
{
case 1:
{
int k=3;
break;
}
case 2:
{
// k is not accessible here
int j;
break;
}
}
}