CODEWITHSUNDEEP
c++struct
Simplicity
Simplicity

Reputation: 48986

C++ - structs... function parameter issue

At: http://www.learncpp.com/cpp-tutorial/82-classes-and-class-members/

It has the following code:

// Declare a DateStruct variable
DateStruct sToday;

// Initialize it manually
sToday.nMonth = 10;
sToday.nDay = 14;
sToday.nYear = 2020;

// Here is a function to initialize a date
void SetDate(DateStruct &sDate, int nMonth, int nDay, int Year)
{
    sDate.nMonth = nMonth;
    sDate.nDay = nDay;
    sDate.nYear = nYear;
}

// Init our date to the same date using the function
SetDate(sToday, 10, 14, 2020);

What is the purpose of the

DateStruct &sDate

parameter in the function signature, especially that I cannot see a use of it in the function body?

Thanks.

Upvotes: 0

Views: 1736

Answers (3)

Marcin Gil
Marcin Gil

Reputation: 69575

The aforementioned, highlighted code is called a reference. You might think of references as aliases for variables.

During function call sDate becomes an alias to sToday - which has been given as a parameter. Thus it makes it possible to modify sToday from within the function!

The reason is that it makes possible to give complex structures that might be filled up with data, modified, etc. inside of called functions.

In your case the SetDate function takes separate year, month and day - and packs it inside sDate (== sToday) structure.

Just compare the first way of initialization (you need to mention all struct members yourself) to calling SetDate function.

eg.:

DateStruct janFirst;
DateStruct decLast;

SetDate(janFirst, 1, 1, 2011);
SetDate(decLast, 12, 31, 2011);

Compare this to how much code would have been written if you had to fill all those janFirst, decLast structures by hand!

Upvotes: 0

Antoine Pelisse
Antoine Pelisse

Reputation: 13129

It means it will take as a first argument a reference to a DateStruct and that this reference will be called sDate in the function's body. The sDate reference is then used in each lines of the body:

sDate.nMonth = nMonth;
sDate.nDay = nDay;
sDate.nYear = nYear;

Upvotes: 4

Puppy
Puppy

Reputation: 147056

It means a reference to an existing DateStruct instance.

Upvotes: 3

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