difference between & and static
closed account (1vf9z8AR)
Is declaring a variable as static or passing it through a function using '&' mean the same thing?
In both cases the original value gets modified.
In both cases the original value gets modified.
no. These things are not related at all!
static variables are never destroyed:
void foo()
{
static int x = 0;
cout << x++ << endl;
}
main()
{
foo(); //writes 0
foo(); //writes 1
foo(); //writes 2...
if you take the static keyword off, you get 0 every time.
& is address of / reference operator. It take an address of a variable:
int x;
int *ip = &x; //ip points to x now.
or as a reference parameter:
void foo(int &x) here, if x is changed inside foo it is changed where it was called from:
{
x = 4;
}
main
int y = 10;
foo(y);
cout << y << endl; //y is 4. foo changed it from &x reference parameter!
static variables are never destroyed:
void foo()
{
static int x = 0;
cout << x++ << endl;
}
main()
{
foo(); //writes 0
foo(); //writes 1
foo(); //writes 2...
if you take the static keyword off, you get 0 every time.
& is address of / reference operator. It take an address of a variable:
int x;
int *ip = &x; //ip points to x now.
or as a reference parameter:
void foo(int &x) here, if x is changed inside foo it is changed where it was called from:
{
x = 4;
}
main
int y = 10;
foo(y);
cout << y << endl; //y is 4. foo changed it from &x reference parameter!
| static variables are never destroyed |
That's a little misleading - objects with static storage duration are destroyed at the end of the program.
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