C中的指针（将地址传递给函数）

Question

I'm trying have a crack at this problem. The question says, *"swap_nums seems to work, but not swap_pointers. Fix it."* I'm a beginner by the way :)

I believe I can work the solution out myself but the trouble is I have a bit of difficulty understanding some programming concepts in C. Here I have shown the given code that requires editing. Below that I will show my thought processes so far. Please note: I want some hints NOT the full solution please. :-)

#include <stdio.h>
#include <stdlib.h>

void swap_nums(int *x, int *y);
void swap_pointers (char *x, char *y);

int main (int argc, char *argv[]){

   int a = 3, b = 4;
   char *s1, *s2;
   swap_nums(&a, &b);
   printf("a is %d\n", a);
   printf("b is %d\n", b);

   s1 = "I should print second";
   s2 = "I should print first";

   swap_pointers(s1, s2);
   printf("s1 is %s\n", s1);
   printf("s2 is %s\n", s2);

   return EXIT_SUCCESS; } 

void swap_nums(int *x, int *y){

   int temp;
   temp = *x;
   *x = *y;
   *y = temp; }

void swap_pointers (char *x, char *y){

   char *temp;
   temp = x;
   x = y;
   y = temp; }

My thought process: This is a program that I believe will supposedly swap the integer variables a and b. It will then take the two declared strings and swap them as well.

Main function:

int a = 3, b = 4;

Assigns variable a and b to 3 and 4 respectively.

char *s1, *s2;

Creates a character pointer variable (which will hold the address of a character)

swap_nums(&a, &b);

The function swap_nums is taking place. I will go to it now to explain my thought processes.

void swap_nums(int *x, int *y){

So I'm not that familiar with passing things into functions, could someone correct what I am about to say if I am wrong here?

The way I see it, we are passing the addresses of a and b as indicated by the ampersand into the function swap_nums. But how come the we have int *x and int *y? I'm a bit confused here. Could someone explain to me this bit?

Answer 1

In C, everything is passed as values, including pointers. Therefore, your code that swaps pointers manipulates pointer copies, leaving originals unchanged.

In order to swap pointers, you need to pass pointers to pointers, not simply pointers. Of course inside the function you need to add a level of dereference the same way that you did in swap_numbers :

void swap_pointers (char **x, char **y) {
    char *temp;
    temp = *x;
    *x = *y;
    *y = temp;
}

Answer 2

A pointer may be an address, but it is still passed by value to functions (like everything in C). To swap two pointers, you need pointers to pointers .

---

Here's what you want:

#include <stdio.h>

int main(void)
{
    char x='s';
    char y='o';
    char *a = &x;
    char *b = &y;
    printf("a is %x\n",a);
    printf("b is %x\n",b);
    printf("swapping\n");
    swap_pointers(&a,&b);
    printf("a is %x\n",a);
    printf("b is %x\n",b);

}

void swap_pointers(char **a, char **b)
{
    char *temp;
    temp = *a;
    *a = *b;
    *b = temp;
}

Answer 3

Here is a hint, there is something wrong with this:

s1 = "I should print second";
s2 = "I should print first";

What is s1 's type, and what is "I should print second"? And how do you save a string literal to a variable?

Answer 4

提示：按值学习调用并通过引用调用以清楚地了解您的查询。

Answer 5

将指针视为另一个变量，我们必须将指向此变量的指针传递给函数swap_pointers() 。

Answer 6

This is what you need i guess. Basic and sufficient.

void swap(char **s1, char **s2){
char *temp=*s1;
*s1=*s2;
*s2=temp;
}

int main(){
char *s1="second";
char *s2="first";
swap(&s1,&s2);
printf("%s",s1);
printf("%s",s2);
return 0;

}  

Hope this helps.