remove a token in c programming

strstr can be used to point to the start of the +instance string (returns, say pointer p0).
   http://www.cplusplus.com/reference/cstring/strstr/

Looks like the ; is your sentinel and you want to keep it.
I assume that ; is not part of the token you want to replace.
Then, you can use strchr to find the ; (returns, say pointer p1).
   http://www.cplusplus.com/reference/cstring/strchr/

strcpy( p0, p1) should copy the ; and beyond over the token starting at p0 - including the terminating null byte.
    http://www.cplusplus.com/reference/cstring/strcpy/

By reviewing the error return values from strstr and strchr, you can add error checking to verify that you have the +instance string token, and the ; that you expect to follow.

It's actually even easier than your last question, and yes, it can be done in the one buffer, e.g.

#include <string.h>

int remove_token(char* buf, const char* token, const char delim) {

    int idx = 0;
    char* delimpos = NULL;
    char* tokenpos = strstr(buf, token);

    if (!tokenpos) return -1; // token not found

    ixd = (int) tokenpos - buf;
    delimpos = strchr(buf + idx,delim);

    if (!delimpos) {
        buf[idx] = '\0';  // no following delimiter, cut of '+instance='

        return strlen(buf); // or just '1', as you like
    }
    
    strcpy(tokenpos, delimpos + 1); // copy the rest to the end of where the token was

    return strlen(buf); // or just '1', as you like
}

int main () {

    char str[] ="<1.2.3.4:4747>;+instance=45;+gks=uuu.";

    remove_token(str, "+instance", ';');
}

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I think there is a bug when the input buffer has a long sentence with carriage return e,g.
<1.2.3.4:4747>;+instance=45;+gks=uuu.\r\n
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx\r\n
yyyyyyyyyyyyyyy\r\n

Etc ...

I m seeing some ovelap at the end , do you know the reason ?

This is for JKr solution.  I will try phoffric solution as well.