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STRUCTURES - eArray[I]=readEmployee...

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STRUCTURES -sometimes need to keep data of different types together -assigning, pass to funct, array elements -can use a structure (structs) struct <structID> { <type1> <ident1>; <type2> <ident2>; …. <typeN> <identN>; }; struct Employee { char lastName[25]; char firstName[25]; double salary; }; int main() { Employee theEmployee; TheEmployee.salary=45000; Strcpy(theEmployee.lastName, “MacLean”); Strcpy(theEmployee.firstName, “james”); Cout<<theEmployee.LastName<<”, “<<theEMployee.firstName<<” $”<<theEmployee.salary<<endl; Employee e1 = theEmployee; Employee e2 = {“Fischer”, “Bobby”, 62000}; Return 0; } Employee readEmployee() { Employee e; Cin>>e.lastName; Cin>>e.firstName; Cin>>e.salary; Return e; }
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Employee eArray[99]; For(int I=0;I<99;I++)
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Unformatted text preview: { eArray[I]=readEmployee(); cout<<eArray[I].salary<<endl; } struct BirthDate { int year; int month; int days; }; struct Employee { char lastName[25]; char firstName[25]; double salary; BirthDate dob; }; Pointers to Structs Employee me = {“Maclean”, “James”, 220000, {1972,07,13}}; Employee *p=&me; Employee *p2 = new Employee; (*p).salary+=20000;// . has higher priority than *-another simpler notation p-> salary += 20000; //works Linked Lists-a “list” is an ordered sequence of elements-a linked list uses pointers to tie the elements together in order-dynamic data structre-grows and shrinks as needed struct ListNode { int data; ListNode *next; }; ListNode *head;...
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STRUCTURES - eArray[I]=readEmployee...

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