AL-zami AL-zami - 1 month ago 18
C Question

Permutation of string

I am trying to implement a code where i have to permute a string.Whenever i try to execute my code i get the following errors and warnings;


1.passing argument 1 of permute makes integer from pointer without a cast

2.expected char but argument is of type char *

3.conflicting type for permute


what might cause those errors in my program?

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

void swap(char *first,char *second);
void permute(char a,int l,int r);

int main(){
char str[] = "ABC";
int size = strlen(str);
permute(str,0,size-1);
return 0;
}
void permute(char *a,int l,int r){
if (l==r){
printf("%s\n",a);
}else{
int i;
for(i=l;i<=r;i++){
swap((a+l),(a+i));
permute(a,l+1,r);
swap((a+l),(a+i));
}
}

}
void swap(char *first,char *second){
char *temp;
*temp = *first;
*first = *second;
*second = *temp;
}

Answer

There's an error in your function prototype. It's declared like this:

void permute(char a,int l,int r);

But defined like this:

void permute(char *a,int l,int r) {

Note that the type of the first argument does not match. You need to change the prototype to match the definition.

Unrelated to that, your swap function is using a pointer temp that is being dereferenced without being set. This is undefined behavior and will likely cause a core dump.

Since you're swapping characters here, you only need a char, not a char *.

void swap(char *first,char *second){
    char temp;
    temp = *first;
    *first = *second;
    *second = temp;
}
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