ISO C++ forbids converting a string constant to ‘char*’ [-Wwrite-strings]

while(client.findUntil("pin", "\n\r"))

This warning indicates that your program is not well-formed. Although you didn’t display the declaration, context can help us to determine that findUntil‘s argument is char*. A string literal may not be passed to such a function.

It was once well-formed, but now deprecated. To pass a string literal such as char* before C++11, it used to work.

I tried:

char const *q = "pin";
char const *r = "\n\r";

These statements are true by themselves.

while(client.findUntil(*q, *r)) 

This is absurd. You were trying to pass a string previously, but you now indirect through the character pointeder so that you are passing a character. This cannot work unless the function is a template.

findUntil(q, r) will not work, either because pointers to const won’t implicitly transform to non-const.

The best solution is to convert string literals into modifiable ranges.

char q[] = "pin";
char r[] = "\n\r";
{
 while(client.findUntil(q, r)) 

Another option is to change findUntil so that it accepts a pointer-to const char instead. String literals can then be used, as they can be converted into a pointer-to const char.