gcc wrapper in C - works (but has unfixable memory leak)

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

/*
This version has an array instead of a string for the command
*/

int main(int argc, char* argv[]) {
    int i;
    int ret = 0;
    int skipped = 0;

    // the new gcc args to pass
    char** gccargs;
    gccargs = (char **)malloc( sizeof(char*) * (argc+1) );
    int gccargc = 0;

    char realcmd[] = "/usr/bin/gcc.real";

    //first pass through args and copy the values if they aren't filtered out
    gccargs[0] = argv[0];
    gccargc++;
    for(i=1; i<argc; i++) {
        char* arg = argv[i];
        if( strcmp(arg, "-Werror") == 0 ) {
            printf("PM gccwrap4: skipping arg: %s\n", arg);
            skipped++;
            continue;
        }
        printf("PM gccwrap4:    \"%s\"\n", arg);

        gccargs[gccargc] = arg;
        gccargc++;
    }
    //next set the next one to 0 so execv knows where to stop
    gccargs[gccargc] = 0;


    printf("PM gccwrap4: final array:\n");
    for(i=0; i<gccargc; i++) {
        printf("PM gccwrap4:     %s\n", gccargs[i]);
    }

    // just for testing; must be commented out because execv does not return if it is successful, so it won't run the real command
//    printf("PM gccwrap4: final command:\n    %s ", realcmd);
//    ret = execv("/bin/echo", gccargs);

    ret = execv(realcmd, gccargs);

    // execv replaces this process, and does not return if successful ... does that mean this free can never be called? does that mean there is no way to free gccargs?
    free(gccargs);

    // don't free gccargs[i] because those are just pointers to argv's values

    return ret;
}