Function that checks Java program arguments -

i'm trying re-write code in cleaner way:

public static void main(string[] args) {     if(checkargs(args) < 0) return;      return; }  private static int checkargs(string[] args) {            if (args.length == 0) {         system.out.println("error: no args");         return -1;     }     try{ //if it's number         int mynumber = integer.parseint(args[0]);         system.out.println("my number is: " + mynumber);     }     catch (exception e) {//if it's string         try{             string[] mystr = args ;             system.out.print("my string ");             (int i=0; i<args.length; i++) {                 mystr[i] = args[i];                 system.out.print(mystr[i]);             }             system.out.print("\n");             return 0;         }         catch(exception err){             system.out.println("error");             return -1;         }     }     return 0; } 

the code checks program args , tell user if it's string or number.

any ideas on how re-write code without using try-catch?

first, don't need second try...catch. code:

        string[] mystr = args ;         system.out.print("my string ");         (int i=0; i<args.length; i++) {             mystr[i] = args[i];             system.out.print(mystr[i]);         }         system.out.print("\n");         return 0; 

is not throwing checked exception , not ever throw runtime exception.

second, code has parts not needed - mystr array - assign args array it, , assign each element individually it. , it's local variable it's going go away return anyway.

so:

        system.out.print("my string ");         (int i=0; i<args.length; i++) {             system.out.print(args[i]);         }         system.out.print("\n");         return 0; 

would same thing , not going throw exceptions.

now, checking integer - think using , catching exception enough. catch numberformatexception instead of exception. it's never idea catch-all.

but if insist on method without try-catch, this:

private static boolean checkint(string str) {     if ( ! str.matches("-?0*[0-9]{1,10}")) {         return false;     }     long l = long.valueof(str);     if ( l > integer.max_value || l < integer.min_value) {         return false;     }     system.out.println("my number is: " + mynumber);     return true; } 

this first checks have no more 10 digits or without minus sign (and number of preceding zeros). if so, can safely converted long, without exception checking. can check if resulting long in range of integer. it's workaround requires no try , catch, don't think it's better try-catch method.

so, try-catch, have:

private static int checkargs1(string[] args) {     if (args == null || args.length == 0) {         system.out.println("error: no args");         return -1;     }     try { // if it's number         int mynumber = integer.parseint(args[0]);         system.out.println("my number is: " + mynumber);     } catch (numberformatexception e) {// if it's string         string[] mystr = args;         system.out.print("my string ");         (int = 0; < args.length; i++) {             mystr[i] = args[i];             system.out.print(mystr[i]);         }         system.out.print("\n");         return 0;     }     return 0; } 

and without it, using checkint() method:

private static int checkargs2(string[] args) {     if (args == null || args.length == 0) {         system.out.println("error: no args");         return -1;     }     if ( ! checkint(args[0])) {         // if it's string         string[] mystr = args;         system.out.print("my string ");         (int = 0; < args.length; i++) {             mystr[i] = args[i];             system.out.print(mystr[i]);         }         system.out.print("\n");         return 0;     }     return 0; } 

i'm not sure how useful check is, though. usually, argument checking done in preparation doing args, , don't save information whether number or string anywhere.