Swapping pointer in C [duplicate]

问题

# include <stdio.h>
void mystery (int *ptra, int *ptrb) {  
     int *temp; 
     temp = ptrb;  
     ptrb =ptra; 
     ptra = temp; 
} 
int main () { 
    int a = 2016, b=0, c= 4, d = 42; 
    mystery (&a, &b);
    if (a < c) 
          mystery (&c, &a); 
    mystery (&a, &d); 
    print f("%d\n", a); 
}

My attempt :

a and d are not swapped as the function mystery() doesn’t change values, but pointers which are local to the function.

Can you explain in formal way, please. how works function mystery?

回答1:

a and d are not swapped as the function mystery() doesn’t change values, but pointers which are local to the function.

Yes, you are right. Function mystery is not swapping values. It is just swapping the pointer ptra and ptrb to location pointed by ptrb and ptra respectively.

                        a
                 +--------------+
                 |              |
ptra ----------> |    2016      |
                 |              |
                 +--------------+



                        b
                 +--------------+
                 |              |
ptrb ----------> |      0       |
                 |              |
                 +--------------+

After executing statements

temp = ptrb;  
ptrb =ptra; 
ptra = temp;    

                        a
                 +--------------+
                 |              |
ptra -----+      |    2016      |
        +------> |              |
        | |      +--------------+
        | |
        | |
        | |
        | |             b
        | |      +--------------+
        | |      |              |
ptrb ---+ +----> |      0       |
                 |              |
                 +--------------+

回答2:

In C, functions always pass by value. So the pointer swapping you're doing has no effect outside the function because you're only changing local values.

Instead of swapping the pointers, you need to swap what they point to, i.e. you need to dereference them:

void mystery (int *ptra, int *ptrb) {  
     const int temp = *ptrb;
     *ptrb = *ptra; 
     *ptra = temp; 
}

回答3:

Your temporary needs to be an int not an int*:

int temp = *ptrb;

Then you need to defererence the pointers: *ptrb = *ptra; etc.

All you're currently doing is swapping the pointers in the function, and this will not be reflected in the function caller.

回答4:

The function mystery() does virtually nothing, and a compiler eliminated almost all of the function via optimization and left only ret.

demo on Compiler Explorer

回答5:

First thing to note: mystery woks on copy of pointers being passeed to it. This has huge implications, most of them being you are not changing original pointers when you assign to pointers ptra, ptrb in your mystery.

void mystery (int *ptra, int *ptrb) {  
     int *temp; 
     temp = ptrb;  
     ptrb =ptra; 
     ptra = temp; 
}

Second thing: you are indeed assigning pointers to pointers, not values of what pointers points to. And given first mistake, the world outside of the mystery sees no changes at all.

Solution:

Dereference pointers to change values they are pointing to.

void
mystery (int *ptra, int *ptrb)
{  
    const int tmp;

    tmp = *ptrb;
    *ptrb = *ptra; 
    *ptra = tmp; 
}

来源：https://stackoverflow.com/questions/35507568/swapping-pointer-in-c

标签

pointers

parameter-passing