computer notes - data structures - 17
TRANSCRIPT
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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
calling function caller
called function intMinus1
stack grows downwards
sp
30
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Reference Variables
Call stack layout after return from intMinus1:
1068
311072
30
callers other stuff
myInt
retVal
calling function caller
stack grows downwards
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
Call stack layout when intMinus2 is called:
1052
1060
1068
1056
311072
?
callers other stuff
myInt
retVal
oldVal1072
calling function caller
called function intMinus2
stack grows downwards
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
calling function caller
called function intMinus2
stack grows downwards
sp
29
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Reference Variables
Call stack layout after return from intMinus2.
1068
311072
29
callers other stuff
myInt
retVal
calling function caller
stack grows downwards
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
Call stack layout when intMinus3 is called:
1052
1060
1068
1056
311072
?
callers other stuff
myInt
retVal
oldVal
calling function caller
called function intMinus3
stack grows downwards
sp
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Reference Variables
Call stack layout when intMinus3 is called:
1052
1060
1068
1056
311072
?
callers other stuff
myInt
retVal
oldVal
calling function caller
called function intMinus3
stack grows downwards
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
calling function caller
called function intMinus3
stack grows downwards
sp
28
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Reference Variables
Call stack layout after return from intMinus3:
1068
311072
28
callers other stuff
myInt
retVal
calling function caller
stack grows downwards
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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