Declared variable within a switch block

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Variable definition inside switch statement 5 answers

Variable Definition Ignore in C [duplicate] 7 answers

Why can't variables be declared in a switch statement? 23 answers

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In this case , it is not undefined behaviour.

For case 0 (when expr == 0 , which is your case), i gets assigned a value before being used (value being read from).

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OK, to elaborate a bit more, for the snippet

switch (expr)
    {
            int i;

            case 0:
                    i = 17;
            default:
                    printf("%dn", i);
    }

just makes the variable i defined in the block scope. Even if, the code would have been written as

            int i = 0; //or any value

the value of i not initialized, it's just the identifier is visible in the scope. You must have another statement assigning value to i before you can make use of it.

In this regard, the C11 standard has a very enlightening example and description. let me quote it, from chapter §6.8.4.2/P7

EXAMPLE In the artificial program fragment

switch (expr)
{
int i = 4;
f(i);
case 0:
i = 17;
/* falls through into default code */
default:
printf("%dn", i);
}

the object whose identifier is ``i exists with automatic storage duration (within the block) but is never initialized, and thus if the controlling expression has a nonzero value, the call to the printf function will access an indeterminate value. [....]

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