C + + pointer parameter assignment problem

Today, there is a problem with the question of assigning a pointer to a function in C + +. The following phenomena can be seen first:

void Test (int *p) {    p = NULL;} int main (int argc, char *argv[]) {    qcoreapplication A (argc, argv);    int *t, y = ten;    t = &y;    Test (t);    return a.exec ();}

　　

I was a little surprised at the result, I always thought that the pointer will be passed, and the pointers are also changed, but the pointer is also a variable, if we want to change it, we must find its address in memory, that is, the address of the pointer. In other words, in a function, if the address of the pointer is assigned, in fact, the original pointer can not be changed!

void Test (int **p) {    *p = NULL;} int main (int argc, char *argv[]) {    qcoreapplication A (argc, argv);    int *t, y = ten;    t = &y;    Test (&t);    return a.exec ();}

　　

In addition, assigning values by reference can also solve this problem:

void Test (int &p) {    int n = 9;    p = N;} int main (int argc, char *argv[]) {    qcoreapplication A (argc, argv);    int T, y = ten;    t = y;    Test (t);    return a.exec ();}

　　

In addition, you can modify what the pointer points to, instead of modifying the pointer address, or you can change the content.

Example 1:

void Test (int *p) {    int n = 9;    *p = n;} int main (int argc, char *argv[]) {    qcoreapplication A (argc, argv);    int *t, y = ten;    t = &y;    Test (t);    return a.exec ();}

　　

Example 2:

void Test (int *p) {    int n = 9;    *p = n;} int main (int argc, char *argv[]) {    qcoreapplication A (argc, argv);    int T, y = ten;    t = y;    Test (&t);    return a.exec ();}

　　

C + + pointer parameter assignment problem