What is the trick for swap_pointers to work?-CodePudding

I am doing an exercise where the task is to fix swap_pointer().

Two string should be swapped.

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

My work so far:

#include <stdio.h>

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

int main(void) {

    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 be printed second.";
    s2 = "I should be printed first.";

    /* swap_pointers(s1, s2); */

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

    return 0;
}

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

    int tmp;

    tmp = *x;
    *x = *y;
    *y = tmp;
}

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

    char tmp;

    tmp = *x;
    *x = *y;
    *y = tmp;
}

I think that not appending & to s1 and s2 before I pass them to the function for swap_pointers is the error?

If anyone can fix my issue here, please don't give the whole solution! I want to use what you give me as a tool to fix it by my own. If I really can't fix it by my own I will adress this later.

Ty in advance!

CodePudding user response：

Let's at first consider this code snippet

int a = 3, b = 4;

swap_nums(&a, &b);

To swap the objects a and b you need to pass then to the function swap_nums by reference through pointers to them. Thus within the function dereferencing the pointers we can get a direct access to original objects and change their values.

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

    int tmp;

    tmp = *x;
    *x = *y;
    *y = tmp;
}

The same we need to do with the variables s1 and s2. That is we need to pass them to the function swap_pointers by reference through pointers to them.

So you need to write

swap_ponters( &s1, &s2 );

and the function will be declared and defined like

void swap_ponters(char **x, char **y)
{
    char *tmp = *x;
    *x = *y;
    *y = tmp;
}

In general if you have two objects like

T a;
T b;

where T is some type specifier then to change the objects in a function you need to pass them to the function by reference through pointers to them. So the declaration of the function swap will look like

void swap( T *a, T *b );

and the function will be called like

swap( &a, &b );

In your program a and b has the type int, that is T is an alias for the type int.

typedef int T;

T a = 3, b = 4;

so the function declaration will look like

  void swap_nums( T *x, T *y );

For the variables s1 and s2 you can write

typedef char *T;

T s1, s2;

and again the function declaration will look like

void swap_pointers( T *x, T *y );

and the function will be called like

swap_pointers( &s1, &s2 );

As in this case T is an alias for the type char * then the function declaration can be rewritten like

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

CodePudding user response：

Your pass the pointers to the first characters of two null-terminated strings to the swap_pointers function.

Inside the function, when you use e.g. *x it's the same as x[0].

So you're swapping the first character of x with the first character of y.

If you want to switch pointers, you need to emulate pass-by-reference like you do with swap_nums, and pass pointers to the pointers, i.e. char **. And use the pointer-to operator & in the call.

So it should be

void swap_ponters(char **x, char **y);

and

swap_pointers(&s1, &s2);