这两种指针方法之间的区别

Question

Decode1:

void decode1(int *xp, int *yp, int *zp){
 int x = *xp;
 int y = *yp; 
 int z = *zp;
 xp = &z;
 yp = &x;
 zp = &y;
}

Decode2:

 void decode2(int *xp, int *yp, int *zp){
 int x = *xp;
 int y = *yp; 
 int z = *zp;
 *xp = z;
 *yp = x;
 *zp = y;
}

Decode1 will change the pointer to address of the z, x, and y. Decode2 will instead change the value at the address of the pointer. Are these two methods interchangeable? Are there situations where one is more correct than the other?

Answer 1

Decode2 is correct procedure. In Decode1, after end of the call stack, the address you have assigned in xp,yp,zp will be vanished.

Answer 2

No they are not interchangeable and they don't do the same thing.

decode1 will not work and decode2 will work (assuming that you want to swap the variables).

The decode1 function places x , y and z on the stack and these variables only exist while the function is executing. The moment it returns these variables no longer point to valid memory. Also, the pointers xp , yp and zp are copies of the pointers that were passed to the function so you won't be modifying the original pointers (hence altering them in the function does absolutely nothing).

The decode2 function will work as expected.

Answer 3

decode1 attempts to returns pointers to local variables but in the end simply does nothing. Use decode2 .