Re: Unexpected pointer action
It displays 10. C++ passes the address of v in main to afunction by
value. The parameter, x, in afunction is the value of the pointer, and
changing it doesn't change v in main.
I think it'll become clearer if you try adding some output to see the
changing values, maybe like so:
void a(int *x) {
std::cout << "a,x " << x << " a,*x " << *x << std::endl;
x = new int(12); // leaks memory
std::cout << "a,x " << x << " a,*x " << *x << std::endl;
}
void testa() {
int v = 10;
std::cout << "testa,&v " << &v << " testa,v " << v << std::=
endl;
a(&v);
std::cout << "testa,&v " << &v << " testa,v " << v << std::=
endl;
}
int main() {
testa();
Thanks for the post, I should know better. If x had not been
reassigned then it would display 12. So I think it would be correct to
say that V was passed by reference (by its address) but it's address
was passed by value.
I had put it many display statements and they were telling me what you
told me but not as concisely and I did not see it.
Thanks again,
jvh (AKA an average programmer)
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