Page 38 - C-Language

problems. Unless you really need them, and these cases are very special, you should avoid them
as you can.

Examples

Character types cannot be accessed through non-character types.

If an object is defined with static, thread, or automatic storage duration, and it has a character
type, either: char, unsigned char, or signed char, it may not be accessed by a non-character type. In
the below example a char array is reinterpreted as the type int, and the behavior is undefined on
every dereference of the int pointer b.

int main( void )
{
char a[100];
int* b = ( int* )&a;
*b = 1;

static char c[100];
b = ( int* )&c;
*b = 2;

_Thread_local char d[100];
b = ( int* )&d;
*b = 3;
}

This is undefined because it violates the "effective type" rule, no data object that has an effective
type may be accessed through another type that is not a character type. Since the other type here
is int, this is not allowed.

Even if alignment and pointer sizes would be known to fit, this would not exempt from this rule,
behavior would still be undefined.

This means in particular that there is no way in standard C to reserve a buffer object of character
type that can be used through pointers with different types, as you would use a buffer that was
received by malloc or similar function.

A correct way to achieve the same goal as in the above example would be to use a union.

typedef union bufType bufType;
union bufType {
char c[sizeof(int[25])];
int i[25];
};

int main( void )
{
bufType a = { .c = { 0 } }; // reserve a buffer and initialize
int* b = a.i; // no cast necessary
*b = 1;

static bufType a = { .c = { 0 } };

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