Replace function won't work with % symbol

Using Arduino Programming Questions
1

I am trying to write a function to replace characters in an array, and it needs to also replace a '%' character with the string: %25

If I am replacing the % character with anything not containing a % symbol, it works, but I won't with the % symbol.

I tried escaping the % with a backslash or another percent symbol. I don't have any idea what is going on.

The input string is: "replace % these % chars %"
and I want it to output: "replace %25 these %25 chars %25"

Code is below, any help is much appreciated.

const int len = 70;
char test[70] = "replace % these % chars  %";

void setup() {
 Serial.begin(9600);
 while (!Serial);
 Serial.println(test);
 replaceChar(test, '%', "hi");
// replaceChar(test, '%', "%25"); //<<this will not work?
 Serial.println(test);
}

void loop() {

}

void replaceChar(char* in, char inchar, char* outstr) {
  char* p = strchr(in, inchar);
  if (p) {

    uint8_t slen = strlen(in);
    uint8_t sublen = strlen(outstr);
    uint8_t a = 0;

    if (sublen > 1) {
      while (p != NULL) {
        if (slen + a < len) {
        memmove(in + (p - in) + (sublen - 1), in + (p - in), slen );
        memcpy(in + (p - in), outstr, sublen);
        p = strchr(p, inchar); 
        a += (sublen - 1);
        }
      }
    }
    else { //if only replacing with one char
      while (p != NULL) {
        in[p - in] = outstr[0];
        p = strchr(p, inchar);
      }
    }
  }
}
2

See if replace letter in a string - C++ Forum helps

Their replaceChar( ) function is more understandable.

3

It might get confused when you replace a '%' with a '%'.

4

KeithRB:
It might get confused when you replace a '%' with a '%'.

It likely gets confused because % becomes %25, which becomes %2525 which becomes %252525, and on and on.

Add some serial print statements to determine what is actually happening. Make sure that you advance the cursor after the substitution.

5

I'm still not sure, the cursor appears to be advancing as it should.
I tried this simpler function:

void replaceChar2(char* in, char inchar){
char* cpystr = "25";
char* p = strchr(in, inchar);

 while (p != NULL) { 
       Serial.println("found");
       Serial.println(in+(p-in));
        memcpy(in+(p-in), cpystr, strlen(cpystr));
        p = strchr(p, inchar);
      }
}

And it still behaves badly only if the search char is '%', and the replace string contains a percent symbol at all. If the copy string has a '%', but the search char is not '%', or vice versa, it works, but when those cases are combined, it breaks.

6

Then time to ditch replaceChar() and write your own.

strchr() and strstr() will help here.

7

These are my own, and they use strchr()

8

what is in + (p-in)? Why not just use p? or just add 1?

9

Yup, in+(p-in) is indeed redundant, I changed it to just p.

In doing that I see that my while loop wasn't correct. as some had suggested it was in the fact that I was updating the pointer to the result of strchr(p, '%') when p itself was just set to '%' to the loop never ended. I changed the pointer update to: p = strchr(p+3, '%') and it worked.

So it had nothing to do with '%' itself, just faulty code.

Thanks for the help.