c - exec() is not throwing an error, but not running the other executable. Syntax error? -

i'm trying use fork() , exec() run program resides in same directery. compiler not complaining, program trying call using execl() not being run. tips? thank you!

pid = fork(); if (pid == -1) {     fprintf (stderr, "error\n");     exit(1);         } else if (pid > 0) {     wait(&status); } else {     execl("./expo.c", "./expo", x, n, (char*) null);     _exit(exit_failure);     } 

i have tried few different versions of exec() , none have worked.

edit: have changed execl("expo","expo",&x,&n,(char*)null); , though still unsure why works based on man pages. man page says first argument should path, not executable. also, why not need ./ second argument if need run executable in terminal?

i have got working correctly, here changed. instead of casting arguments (x , n) ints passing them child, passed them chars , casted them ints in child process.

if(pid==-1){         fprintf (stderr, "error\n");     exit(1);             }     else if(pid>0){         wait(&status);     }     else{         const char *x=argv[1];         const char *n=argv[2];          execl("expo","expo",x,n,(char*)null);         perror("execl() failure!\n");                exit(exit_failure);          }