(Beginner) Don't know why code doesn't give correct output

Question

Firstly, I'm trying to create a program which takes two char arrays and removes from the first any chars that shew up in the second array. I have added comments to make it easier to understand what's happening.

   #include <stdio.h>

void rcfs(char s[], char t[]){
    int i=0, j=0;
    while (t[i] != '\0')
        ++i; /*find size of t*/
    for (int x=0;x<i;++x){ /*go through both vals of x and
    check vals of s to see if they equal a val of x, if so set j of s to 0*/
        while (s[j]!=0){
            if (s[j]==t[x]){
                s[j]=0;
            }
            j++;
        }
        if (x<i-1)
            j=0;
    }
    char new[8]; /*new char arr to be assigned to s*/
    int x=0; //seperate counter that increments only when chars are added to new
    for (int q=0;q<j;q++){
        if (s[q]!=0)
            new[x++]=s[q];
    }
    new[7]=0;
    s=new; //maybe these need be pointers
}

int main(){
    char s[]="I eat food";
    char t[]="oe";
    printf("Pre: %s\n", s);
    rcfs(s, t);
    printf("Post: %s", s);
    return 0;
}

Here is the output: Instead of 'Post: I' it should be: 'Post: I at fd' To summarise, I'm getting the wrong output and I'm currently unaware of how to fix it.

Answer 1

I suggest that you turn the algorithm around and loop over s in the outer loop. I also suggest that you use strlen to find the length of strings and strchr to find out if a character is present in a string.

Example:

#include <string.h>

void rcfs(char *s, char t[]) {   
    char *sw = s; // current writing position

    for(; *s != '\0'; ++s)  {        // s is the current reading pos
        if(strchr(t, *s) == NULL) {  // *s is not in t
            *sw = *s;                // save this character at sw
            ++sw;                    // and step sw
        }        
    }

    *sw = '\0'; // null terminate
}

Demo

Answer 2

Your approach is entirely wrong and erroneous.

For example this while loop

    while (s[j]!=0){
        if (s[j]==t[x]){
            s[j]=0;
        }
        j++;
    }

can be unable to find a character equal to the character t[x] if already within the string s you set some preceding character to 0.

For example if s = "12" and t = "12" then after the first iteration of the for loop

for (int x=0;x<i;++x){

the array s will look like s = "\02" So the while loop in the second iteration of the for loop will exit at once due to the condition

while (s[j]!=0){

This declaration of a character array with the magic number 8

char new[8];

does not make a sense. The user can pass to the function strings of any length.

In this assignment statement

s=new

the local variable s of the pointer type char * (due to adjusting by the compiler the parameter having an array type to pointer to the array element type) is assigned by the address of the first element of the local array new. This does not change the source array s declared in main.

Without using standard C string functions your function can be declared and defined the following way as it is shown in the demonstration program below.

#include <stdio.h>

char *rcfs( char s1[], const char s2[] )
{
    if (*s2)
    {
        char *src = s1, *dsn = s1;
        do
        {
            size_t i = 0;
            while (s2[i] != '\0' && s2[i] != *src) ++i;
            if (s2[i] == '\0')
            {
                if (src != dsn) *dsn = *src;
                ++dsn;
            }
        } while (*src++);
    }

    return s1;
}

int main( void )
{
    char s[] = "I eat food";

    printf( "Pre: %s\n", s );

    printf( "Post: %s\n", rcfs(s, "oe"));
}

The program output is

Pre: I eat food
Post: I at fd

Answer 3

To begin with:

Instantiate char new[8] outside of function rcfs, pass it to that function, and print it after the function returns:

char new[8];
rcfs(s, t, new);
printf("Post: %s", new);

The function should be declared as void rcfs(char s[], char t[], char new[]) of course.