Swapping two arrays using pointer

Question

i I was wondering why code below doesn't work. I have used pointer to copy array from one to another but it doesn't copy at all. Am I missing something?

#include <stdio.h> 
typedef struct student {
    int id;
    char *pname;
    double points;
} STUD;

void stud_printx(STUD s) {
    printf("[%d:%s] = %lf\n", s.id, s.pname, s.points);
}

void stud_swap(STUD *s1, STUD *s2) { // space to be filled - my code written
STUD tmp;
     tmp = *s1;
     *s1 = *s2; 
     *s1 = tmp;    


}

int main(void) {
    STUD s1 = {1, "Choi", 9.9};
    STUD s2 = {2, "Park", 0.1};

    stud_printx(s1);
    stud_printx(s2);

    stud_swap(&s1, &s2 ); // space to be filled  - my code written 

    stud_printx(s1);
    stud_printx(s2);

    return 0;
}

Re "I have used pointer to copy array", huh, there's no array in that program. — ikegami
– ikegami, Commented Nov 21, 2017 at 1:57
Voting to close as typo - it seems clear that you had *s2 = tmp; in mind but just didn't type it out right — M.M
– M.M, Commented Nov 21, 2017 at 3:53

ikegami · Accepted Answer · 2017-11-21 02:00:35Z

2

*s1 = *s2;  // Copy original *s2 into *s1
*s1 = tmp;  // Copy original *s1 into *s1

should be

*s1 = *s2;  // Copy original *s2 into *s1
*s2 = tmp;  // Copy original *s1 into *s2

answered Nov 21, 2017 at 2:00

ikegami

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1 Comment

Nathan Lee Over a year ago

I've looked over the code again and just noticed the mistake. Thank you!!

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