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;
}
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
}
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);
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