Question

I want to know what’s wrong with my code.
It’s supposed to exchange the two word specified by the user,
but it does nothing. A little help would be nice.

#include <stdio.h>

void changeW(char *ptab1, char *ptab2){
char tmp;
tmp = *ptab1;
*ptab1 = *ptab2;
*ptab2 = tmp;
printf("%s %s",ptab1,ptab2);
return;
}
int main(void) {

char tab1[25];
char tab2[25];

printf("type two words");
scanf("%s %s",tab1,tab2);
changeW(tab1,tab2);
return 0;
}

corrected code, but still a problem ! i can swap small words but when they get long, i got weird characters in the terminal such as �����.

void changeW(char *ptab1, char *ptab2){
int l;
if(length(ptab1)<length(ptab2)){
l = length(ptab2);
}
else {l=length(ptab1);}
for(int i=0; i<l;i++){
char tmp;
tmp =ptab1[i];
ptab1[i] =ptab2[i];
ptab2[i]=tmp;
}
printf("%s %s",ptab1,ptab2);
return;
}
int main(void) {

char tab1[25];
char tab2[25];

printf("type two words");
scanf("%s %s",tab1,tab2);
changeW(tab1,tab2);
return 0;
}

Ok i found the solution thanks all for your help.
All you got to do is, in changeW,

printf("%s\t%s",ptab1,ptab2);

the single space seems not enough to separate both words, a tab is fine.

Last edit :
in fact, searching for the longest table is useless, as tab1 and tab2 are both 25 character long.

for(int i=0; i<25;i++)

works fine.

Answer 1

void changeW(char *ptab1, char *ptab2)
{
    size_t l1 = strlen(ptab1), l2 = strlen(ptab2);
    size_t l = l1 > l2 ? l1 : l2;
    size_t i;
    for (i = 0; i <= l; i++) {
        char tmp = ptab1[i];
        ptab1[i] = ptab2[i];
        ptab2[i] = tmp;
    }
}

To exchange the two words without a temporary buffer variable you’ll need to do some bit tricks.

void changeW(char *ptab1, char *ptab2)
{
    size_t l1 = strlen(ptab1), l2 = strlen(ptab2);
    size_t l = l1 > l2 ? l1 : l2;
    size_t i;
    for (i = 0; i <= l; i++) {
        ptab1[i] ^= ptab2[i];
        ptab2[i] ^= ptab1[i];
        ptab1[i] ^= ptab2[i];
    }
}

The ^ operator is the bitwise XOR boolean operation.

0^0=0
0^1=1
1^0=1
1^1=0

The trick that makes this work is that (a^b) ^ a=b and b^(a^b)=a for any a and b. The swapping part is thus:

new_p2 = p2 ^ (p1 ^ p2) = p1
new_p1 = (p1 ^ p2) ^ new_p2 = (p1 ^ p2) ^ p1 = p2