Question

while (1)
{
    char j;
    if (x[j] == y[j])

Here I am trying to start a loop where I want to able to match any of the characters from char array ‘x’ with char array ‘y’. If the characters do match from x to y then I want to keep them as they are and if they don’t I want to be able to replace them with a star ‘*’. (e.i. x = [a,p,f] and y = [a,p,p,l,e] and so after the match up and replacement y = [a,p,p,*,*] and when I cout it spells out app**)

I have no idea how to set this up and what type of loop I should use. I fairly new to programming and I know basic replace and switch functions.

Answer 1

This more or less does what you specify, I think.

#include <string.h>

for (int j = 0; y[j] != '\0'; j++)
{
    if (strchr(x, y[j]) == 0)
        y[j] = '*';
}

---

Test program

@LooneyTunes asks what happens with: x[] = "apcd" and y[] = "abcd" – do you get "a*cd".
The answer is yes. Here’s a test program that demonstrates the results. As far as I am concerned, it is pure C code, though G++ is quite happy with it too. You might need the C99 option such as ‘-std=c99‘ with GCC set on the compiler. MSVC won’t like it if it compiles this as C code; declare j at the top of the function for it.

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

static void doit(const char *x, char *y)
{
    printf("Before: x = %s, y = %s\n", x, y);
    for (int j = 0; y[j] != '\0'; j++)
    {
        if (strchr(x, y[j]) == 0)
            y[j] = '*';
    }
    printf("After:  x = %s, y = %s\n", x, y);
}

int main(void)
{
    const char x1[] = "apf";
    const char x2[] = "apcd";
    char       y1[] = "apple";
    char       y2[] = "abcd";

    doit(x1, y1);
    doit(x2, y2);
    return 0;
}

Output

Before: x = apf, y = apple
After:  x = apf, y = app**
Before: x = apcd, y = abcd
After:  x = apcd, y = a*cd