Computer Notes - Data Structures - 17

8/3/2019 Computer Notes - Data Structures - 17

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Class No.17

Data Structures

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Reference Variables

The symbol & has a few different purposesdepending on where it occurs in code.

When it appears in front of a variable name, itis the address operator, i.e., it returns theaddress of the variable in memory.

int x;int* ptr = &x;

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Reference Variables

The symbol & can also appear after a typein a function signature:

For example, insert and remove from theBinarySearchTree class.

void insert( const EType& x );

void remove( const EType& x );

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Reference Variables

Or, in case we designed theBinarySearchTree class to hold integers

only, i.e., no templates

void insert( const int& x );

void remove( const int& x );

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Reference Variables

The & indicates a parameter that is a

reference variable.

Consider the following three differentfunctions:

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Reference Variables

// example 1

int intMinus1( int oldVal)

{

oldVal = oldVal 1;

return oldVal;

}



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Reference Variables

// example 2

int intMinus2( int* oldVal)

{

*oldVal = *oldVal 2;

return *oldVal;

}



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Reference Variables

// example 3

int intMinus3( int& oldVal)

{

oldVal = oldVal 3;

return oldVal;

}



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Reference Variables

The caller function: calling intMinus1

void caller(){

int myInt = 31;

int retVal;

retVal = intMinus1( myInt );

cout


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Memory Organization

Code

Static data

Stack

Heap

Process 1(browser)

Process 3(word)

Process 4(ourtest.exe)

Windows OS

Process 2(dev-c++)

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Reference Variables

Call stack layout

Parameters(caller)

Local variables(caller)

Return address(caller)

Parameters(intMinus1)

Local variables(intMinus1)

Return address(intMinus1)sp

Stack grows downwards

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Reference Variables

Call stack layout when intMinus1 is called:

1052

1060

1068

1056

311072

?

callers other stuff

myInt

retVal

oldVal31

calling function caller

called function intMinus1

stack grows downwards

sp
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Reference Variables

How are myInt and oldVal related?

Passing myInt to the functionintMinus1 results in a copyof myIntto be placed in parameter oldVal inthe call stack.

Alterations are done to the copy of 31(stored in oldVal) and not the originalmyInt.

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Reference Variables

The original myInt remainsunchanged.

For this reason, this technique ofpassing parameters is called pass byvalue.

Alterations are done to the copy of 31(stored in oldVal) and not the originalmyInt.



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Reference Variables

Call stack layout after subtraction in intMinus1:

1052

1060

1068

1056

311072

?

callers other stuff

myInt

retVal

oldVal31




sp

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Reference Variables

Call stack layout after return from intMinus1:

1068

311072

30

callers other stuff

myInt

retVal



sp


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Reference Variables

We could have called intMinus1 asvoid caller()

{

int retVal;

retVal = intMinus1( 31 );// literal

cout


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Reference Variables

If the programmer wanted to actually change avariables value from within a function, one waywould be to send a pointer:

void caller(){

int retVal;

int myInt = 31;

retVal = intMinus2( &myInt );cout


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Reference Variables


1052

1060

1068

1056

311072

?

callers other stuff

myInt

retVal

oldVal1072




sp


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Reference Variables

Call stack layout after *oldVal = *oldVal 2;

1052

1060

1068

1056

311072

?

callers other stuff

myInt

retVal

oldVal1072




sp

29


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Reference Variables

Call stack layout after return from intMinus2.

1068

311072

29

callers other stuff

myInt

retVal



sp

29


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Reference Variables

Suppose we want a function to changean object.

But we dont want to send the function acopy. The object could be large andcopying it costs time.

We dont want to use pointers becauseof the messy syntax.

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Reference Variables

The answer: call-by-reference(or pass-by-reference):

void caller(){

int retVal;

int myInt = 31;retVal =

intMinus3( myInt );

cout


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Reference Variables

The & after int means that oldVal is aninteger reference variable.

// example 3

int intMinus3( int& oldVal)

{

oldVal = oldVal 3;

return oldVal;

}



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Reference Variables

So what isa reference variable?

The idea is: the integer object myInt is used

exactly as it exists in the caller. The functionsimply reaches it through a different name,oldVal.

The function intMinus3 cannot use the name

myInt because it is in the callers scope. But both variable names refer to the same

object (same memory cell).

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Reference Variables


1052

1060

1068

1056

311072

?

callers other stuff

myInt

retVal

oldVal




sp
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Reference Variables


1052

1060

1068

1056

311072

?

callers other stuff

myInt

retVal

oldVal




sp
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Reference Variables

Call stack layout after oldVal = oldVal - 3:

1052

1060

1068

1056

311072

?

callers other stuff

myInt

retVal

oldVal




sp

28


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Reference Variables

Call stack layout after return from intMinus3:

1068

311072

28

callers other stuff

myInt

retVal



sp

28
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Reference Variables

The compiler may actually implement call-by-reference using pointers as we did in

example 2. The obtaining of address and de-

referencing would be done behind thescene.

We should think in terms of therenaming abstraction.

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Computer Notes - Data Structures - 17

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