Swap char with table pointers in C

Question

I'm trying to swap two char with two table pointers. Can someone explain to me what's wrong in my code? The terminal says char** is expected but I don't know what to do, so I think I don't really understand how pointers work for tables.

void echangeM2(char **ptab1, char **ptab2){

  char *tmp = *ptab1;
  *ptab1 = *ptab2;
  *ptab2 = *tmp;
  printf("%s\t %s",*ptab1,*ptab2);

  return;
}

int main(void) {
  char tab1[25];
  char tab2[25];
  char *adtab1;
  char *adtab2;
  *adtab1 = &tab1;
  *adtab2=&tab2;
  printf("type two words");
  scanf("%s %s",tab1,tab2);
  echangeM2(adtab1,adtab2);
  return 0;
}

Check this [link](http://stackoverflow.com/questions/4955198/what-does-dereferencing-a-pointer-mean) which helps understanding reference and deference in context of pointers in `C` — Sunil Bojanapally, Dec 16 '12 at 16:08

Masked Man · Accepted Answer · 2012-12-16T15:58:01.990

1

The following code should work for you:

#include <stdio.h>

void exchangeM2(char* *ptab1, char* *ptab2) { // accepts pointer to char*
  char* tmp = *ptab1;  // ptab1's "pointed to" is assigned to tmp
  *ptab1 = *ptab2;     // move ptab2's "pointed to" to ptab1
  *ptab2 = tmp;        // now move tmp to ptab2
  printf("%s\t %s",*ptab1,*ptab2);
}

int main(void) {
  char tab1[25];
  char tab2[25];
  char* adtab1;
  char* adtab2;
  adtab1 = tab1;  // array name itself can be used as pointer
  adtab2 = tab2;
  printf("type two words");
  scanf("%s %s",tab1,tab2);
  exchangeM2(&adtab1, &adtab2);  // pass the address of the pointers to the function
}

edited Dec 16 '12 at 15:58

answered Dec 16 '12 at 15:51

Masked Man

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i'd really like but I only have 10 reputation xD – Dany77 Dec 17 '12 at 09:59
Ah sorry, I did not realise that. – Masked Man Dec 17 '12 at 11:31

score 0 · Answer 2 · answered Dec 16 '12 at 15:50

echangeM2(&adtab1,&adtab2);

This should fix the compile errors. You are passing char* pointers to a function that expects a char ** pointer

Edit: Actually looks like you want something like

char **adtab1;
char **adtab2;
adtab1 = &tab1;
adtab2=&tab2;
...
echangeM2(adtab1,adtab2);

Swap char with table pointers in C

2 Answers2