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Presbiopia - la 43 ani ?

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Casti USB-C ptr A-54

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 ERR_ADDRESS_UNREACHABLE

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cum pot conditiona trecerea la urmatoarea secventa

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2 replies to this topic

#1
legalizes_crime

legalizes_crime

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  • Grup: Members
  • Posts: 761
  • Înscris: 13.08.2006
int main() {
   char last_name[20], first_name[30], cnp[14], phone[11], email[40];
   cout << " please input your last name " << endl;
   cin.getline(last_name, 19);
   cin.ignore();
   //int character_check(char last_name[]);
   if(('A' <= last_name[i] && last_name[i] <= 'Z') || ('a' <= last_name[i] && last_name[i] <= 'z')) {
		for(int i=0; i<strlen(last_name); i++)
		last_name[i] = toupper(last_name[i]);
		cout << " your name is " << last_name << endl;
	} else {
		cout << " please characters only ";


   /////. cum fortez utilizatorul sa introduca doar caractere,iar,in caz contrar, sa nu treaca codul mai departe?  ar trebui cu do...while. cred ?




	}
   cout << " please enter your first name " << endl;
   cin.getline(first_name, 29);
   cin.ignore();
   first_name[0] = toupper(first_name[0]);
   for(i=1;i<strlen(first_name);i++)
   if(first_name[i] == ' ')
   first_name[i+1] = toupper(first_name[i+1]);
   cout << " your first name is " << first_name << endl;


Edited by MarianG, 13 July 2022 - 09:31.
tag-uri code


#2
MarianG

MarianG

    be that as it may

  • Grup: Moderators
  • Posts: 31,428
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https://stackoverflo...nction#32911580
bool is_valid_input(std::string const& name)
{
   bool is_valid = true;
   // Figure out the logic for deciding when the input is not valid.
   // ...
   if (!is_valid )
   {
	  cout << "Numbers are not allowed, input name again: ";
   }
   return is_valid;
}
int main()
{
	string name = "";
	do
	{
	   cout << "Enter name: ";
	   getline(cin, name);
	}
	while ( !is_valid_input(name) );
}


#3
legalizes_crime

legalizes_crime

    Member

  • Grup: Members
  • Posts: 761
  • Înscris: 13.08.2006
multumesc mult pentru ajutor.

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