By running this code
char array[6];
int i;
for ( i = 0; i < 6; ++i )
printf("%i ", array[i]);
Possible output:
64 0 -64 77 67 0
I get always the last element 0, although I was expecting random value. It is compiler dependent? I'm using gcc.
No. There's no such thing guaranteed by the C standard for local variables.
The values of the uninitialized array has indeterminate values. So, you can't access them and since you do, your code has undefined behaviour.
But the variables with static storage duration such as global variables, static qualified variables etc are initialized with zero.
The contents of a variable (and by extension, the elements of an array) that do not have static storage duration (globals, static locals) are undefined.
The fact that the last element in the array happens to be 0 essentially is random.
Last element is zero in string constants like "Test" or char array[] = "Test";. In your example last value is zero by chance.
Try this:
void f1() // prepare non-zero stack
{
char array[40];
memset( array, 32, sizeof array );
}
void f2() // your array
{
char array[6];
int i;
for ( i = 0; i < 6; ++i )
printf("%i ", array[i]);
}
int main()
{
f1();
f2();
return 0;
}
