1
Workin’ with PointasWorkin’ with Pointas
An exercise in destroying your computer
2
What is this?What is this?
• Your worst nightmare!• Comes from pointer misuse
3
Let’s look at Memory!Let’s look at Memory!Blue is memory address, Blue is memory address, Black is value, Black is value, Red is variable nameRed is variable name
1 -4717
2-901
3 76
4-0
5 98131
6 -1038
7 -554
8 7462
9 312
10 -6
11 3619
12 -4717
13 60981
14 4148
15 86851
16 -5155
17 95151
18 -47
19 2251
20 0
21 -78781
22 -901
23-6
24 6720
25 -4717
26 -19
2721511
28 -9
29 17
30 -6561
31 -651
32 9
33 761
34 -896761
35 7851
36 -6
37 9996
38 674547
39 -6868
40 -1
41 5431
42 -4717
4
Declare an intDeclare an intint myInt;int myInt;
1 -4717
2-901
3 76
4-0
5 98131
6 -1038
7 -554
8 7462
9 312
10 -6
11 3619
12 -4717
13 60981
14 4148
15 86851
16 -5155
17 95151
18 -47
19 2251
20 0
21 -78781
22 -901
23-6
24 6720
25 myInt-4717
26 -19
2721511
28 -9
29 17
30 -6561
31 -651
32 9
33 761
34 -896761
35 7851
36 -6
37 9996
38 674547
39 -6868
40 -1
41 5431
42 -4717
5
What’ve we done?What’ve we done?
• By declaring the int, we’ve taken up just enough memory to hold an int
• We don’t know where in memory (the address) that it’s located
• Computer picks “at random”• What value is at that memory
location?• Can we print that value out?
– The value –4717 would print! (garbage)
6
Copy 42 into that Section of Copy 42 into that Section of MemoryMemory
myInt = 42;myInt = 42;1 -4717
2-901
3 76
4-0
5 98131
6 -1038
7 -554
8 7462
9 312
10 -6
11 3619
12 -4717
13 60981
14 4148
15 86851
16 -5155
17 95151
18 -47
19 2251
20 0
21 -78781
22 -901
23-6
24 6720
25 myInt 26 -19
2721511
28 -9
29 17
30 -6561
31 -651
32 9
33 761
34 -896761
35 7851
36 -6
37 9996
38 674547
39 -6868
40 -1
41 5431
42 -4717
42
7
PointersPointers
• Allow us to get to the address of where information is located
• Similar to call forwarding– Ask the pointer where to go– Go there for the information
• To create a pointer, we use the *• Still follows format of <data type>
<name>;• Example:
int* ptr;
8
Declare an int pointerDeclare an int pointerint* ptr;int* ptr;
1 -4717
2-901
3 76
4-0
5 ptr98131
6 -1038
7 -554
8 7462
9 312
10 -6
11 3619
12 -4717
13 60981
14 4148
15 86851
16 -5155
17 95151
18 -47
19 2251
20 0
21 -78781
22 -901
23-6
24 6720
25 myInt42
26 -19
2721511
28 -9
29 17
30 -6561
31 -651
32 9
33 761
34 -896761
35 7851
36 -6
37 9996
38 674547
39 -6868
40 -1
41 5431
42 -4717
9
Now what have we done?Now what have we done?
• Created a new variable that’s of type ptr to a int
• Notice that we haven’t initialized the pointer to “point” to myInt yet
• What if we print the pointer out?
10
cout << ptr;cout << ptr;(prints out value of ptr: 98131)(prints out value of ptr: 98131)
1 -4717
2-901
3 76
4-0
5 ptr98131
6 -1038
7 -554
8 7462
9 312
10 -6
11 3619
12 -4717
13 60981
14 4148
15 86851
16 -5155
17 95151
18 -47
19 2251
20 0
21 -78781
22 -901
23-6
24 6720
25 myInt42
26 -19
2721511
28 -9
29 17
30 -6561
31 -651
32 9
33 761
34 -896761
35 7851
36 -6
37 9996
38 674547
39 -6868
40 -1
41 5431
42 -4717
11
ProblemProblem
• How do we get address of myInt so ptr can point to it?
• Remember, we can still access the value of myInt directlyint someInt = myInt;
• We really need the pointer to store the address of where myInt is located
• We do not need to store the value of myInt for the pointer (just the address)
12
The & operatorThe & operator
• Use the & operator to get the address of where the variable is in memory
• What would the following statement print to the screen?cout << &myInt << endl;
13
What would happen?What would happen?cout << &myInt;cout << &myInt;
1 -4717
2-901
3 76
4-0
5 ptr98131
6 -1038
7 -554
8 7462
9 312
10 -6
11 3619
12 -4717
13 60981
14 4148
15 86851
16 -5155
17 95151
18 -47
19 2251
20 0
21 -78781
22 -901
23-6
24 6720
25 myInt42
26 -19
2721511
28 -9
29 17
30 -6561
31 -651
32 9
33 761
34 -896761
35 7851
36 -6
37 9996
38 674547
39 -6868
40 -1
41 5431
42 -4717
14
Getting the Pointer to PointGetting the Pointer to Point
• We now need “ptr” to point to myInt
• Code:ptr = &myInt;
ptr is a pointer,so it expects anaddress to be assigned to it
Here, we get the address of wheremyInt is stored in memory and copythat value into “ptr”
15
BeforeBefore
1 -4717
2-901
3 76
4-0
5 ptr98131
6 -1038
7 -554
8 7462
9 312
10 -6
11 3619
12 -4717
13 60981
14 4148
15 86851
16 -5155
17 95151
18 -47
19 2251
20 0
21 -78781
22 -901
23-6
24 6720
25 myInt42
26 -19
2721511
28 -9
29 17
30 -6561
31 -651
32 9
33 761
34 -896761
35 7851
36 -6
37 9996
38 674547
39 -6868
40 -1
41 5431
42 -4717
16
AfterAfterptr = &myInt;ptr = &myInt;
1 -4717
2-901
3 76
4-0
5 ptr25
6 -1038
7 -554
8 7462
9 312
10 -6
11 3619
12 -4717
13 60981
14 4148
15 86851
16 -5155
17 95151
18 -47
19 2251
20 0
21 -78781
22 -901
23-6
24 6720
25 myInt42
26 -19
2721511
28 -9
29 17
30 -6561
31 -651
32 9
33 761
34 -896761
35 7851
36 -6
37 9996
38 674547
39 -6868
40 -1
41 5431
42 -4717
17
What would this do?What would this do?ptr = myInt;ptr = myInt;
1 -4717
2-901
3 76
4-0
5 ptr98186
6 -1038
7 -554
8 7462
9 312
10 -6
11 3619
12 -4717
13 60981
14 4148
15 86851
16 -5155
17 95151
18 -47
19 2251
20 0
21 -78781
22 -901
23-6
24 6720
25 myInt42
26 -19
2721511
28 -9
29 17
30 -6561
31 -651
32 9
33 761
34 -896761
35 7851
36 -6
37 9996
38 674547
39 -6868
40 -1
41 5431
42 -4717
18
Oh no!Oh no!ptr = myInt;ptr = myInt;
1 -4717
2-901
3 76
4-0
5 ptr42
6 -1038
7 -554
8 7462
9 312
10 -6
11 3619
12 -4717
13 60981
14 4148
15 86851
16 -5155
17 95151
18 -47
19 2251
20 0
21 -78781
22 -901
23-6
24 6720
25 myInt42
26 -19
2721511
28 -9
29 17
30 -6561
31 -651
32 9
33 761
34 -896761
35 7851
36 -6
37 9996
38 674547
39 -6868
40 -1
41 5431
42 -4717
19
Tricky Screens of Death!Tricky Screens of Death!
• Last thing to learn is how to “dereference” pointer
• This means “how to follow the pointer”
• Unfortunately, we use the * operator as well
• Example:cout << *ptr << endl; //Follow wherever ptr is pointing to and print that value out!
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Follow the Pointer and Print Follow the Pointer and Print it Outit Out
cout << *ptr << endl;cout << *ptr << endl;1 -4717
2-901
3 76
4-0
5 ptr25
6 -1038
7 -554
8 7462
9 312
10 -6
11 3619
12 -4717
13 60981
14 4148
15 86851
16 -5155
17 95151
18 -47
19 2251
20 0
21 -78781
22 -901
23-6
24 6720
25 myInt42
26 -19
2721511
28 -9
29 17
30 -6561
31 -651
32 9
33 761
34 -896761
35 7851
36 -6
37 9996
38 674547
39 -6868
40 -1
41 5431
42 -4717
21
Another ExampleAnother ExampleBlue is memory address, Blue is memory address, Black is value, Black is value, Red is variable nameRed is variable name
1 -4717
2-901
3 76
4-0
5 98131
6 -1038
7 -554
8 7462
9 312
10 -6
11 3619
12 -4717
13 60981
14 4148
15 86851
16 -5155
17 95151
18 -47
19 2251
20 0
21 -78781
22 -901
23-6
24 6720
25 -4717
26 -19
2721511
28 -9
29 17
30 -6561
31 -651
32 9
33 761
34 -896761
35 7851
36 -6
37 9996
38 674547
39 -6868
40 -1
41 5431
42 -4717
22
Declare a PointerDeclare a Pointerint *ptr;int *ptr;
1 -4717
2-901
3 76
4-0
5 98131
6 -1038
7 -554
8 7462
9 312
10 ptr -6
11 3619
12 -4717
13 60981
14 4148
15 86851
16 -5155
17 95151
18 -47
19 2251
20 0
21 -78781
22 -901
23-6
24 6720
25 -4717
26 -19
2721511
28 -9
29 17
30 -6561
31 -651
32 9
33 761
34 -896761
35 7851
36 -6
37 9996
38 674547
39 -6868
40 -1
41 5431
42 -4717
23
What would happen?What would happen?cout << *ptr << endl;cout << *ptr << endl;
1 -4717
2-901
3 76
4-0
5 98131
6 -1038
7 -554
8 7462
9 312
10 ptr -6
11 3619
12 -4717
13 60981
14 4148
15 86851
16 -5155
17 95151
18 -47
19 2251
20 0
21 -78781
22 -901
23-6
24 6720
25 -4717
26 -19
2721511
28 -9
29 17
30 -6561
31 -651
32 9
33 761
34 -896761
35 7851
36 -6
37 9996
38 674547
39 -6868
40 -1
41 5431
42 -4717
24
BSOD!BSOD!
25
• Because parameter passing only passes a copy so the function can’t change main’s variables!void cannotChange (int x) {
x = 6;cout << x << endl;
}void main ( ) {
int myInt = 17;cannotChange (myInt);cout << myInt << endl;
}
Why do I need Pointers?Why do I need Pointers?
0 1 2
3 4 5
6 7 8
-2 91 571
-2991 0 -33
41 61 -1
26
• Because parameter passing only passes a copy so the function can’t change main’s variables!void cannotChange (int x) {
x = 6;cout << x << endl;
}void main ( ) {
int myInt = 17;cannotChange (myInt);cout << myInt << endl;
}
Declare myIntDeclare myInt
0 1 2
3 4 5
6 7 8
-2 91 571
-2991 0 -33
41 17 -1
myInt
27
• Because parameter passing only passes a copy so the function can’t change main’s variables!void cannotChange (int x) {
x = 6;cout << x << endl;
}void main ( ) {
int myInt = 17;cannotChange (myInt);cout << myInt << endl;
}
Call the functionCall the function
0 1 2
3 4 5
6 7 8
-2 91 571
-2991 0 -33
41 17 -1
myInt
28
• Because parameter passing only passes a copy so the function can’t change main’s variables!void cannotChange (int x) {
x = 6;cout << x << endl;
}void main ( ) {
int myInt = 17;cannotChange (myInt);cout << myInt << endl;
}
Here’s where the Copy is Here’s where the Copy is MadeMade
0 1 2
3 4 5
6 7 8
-2 17 571
-2991 0 -33
41 17 -1
myInt
x
29
• Because parameter passing only passes a copy so the function can’t change main’s variables!void cannotChange (int x) {
x = 6;cout << x << endl;
}void main ( ) {
int myInt = 17;cannotChange (myInt);cout << myInt << endl;
}
Changing Only Local CopyChanging Only Local Copy
0 1 2
3 4 5
6 7 8
-2 6 571
-2991 0 -33
41 17 -1
myInt
x
30
• Because parameter passing only passes a copy so the function can’t change main’s variables!void cannotChange (int x) {
x = 6;cout << x << endl;
}void main ( ) {
int myInt = 17;cannotChange (myInt);cout << myInt << endl;
}
Print Out Local Copy (6)Print Out Local Copy (6)
0 1 2
3 4 5
6 7 8
-2 6 571
-2991 0 -33
41 17 -1
myInt
x
31
• Because parameter passing only passes a copy so the function can’t change main’s variables!void cannotChange (int x) {
x = 6;cout << x << endl;
}void main ( ) {
int myInt = 17;cannotChange (myInt);cout << myInt << endl;
}
Return to Main (print 17)Return to Main (print 17)(x is gone and leaves garbage)(x is gone and leaves garbage)
0 1 2
3 4 5
6 7 8
-2 6 571
-2991 0 -33
41 17 -1
myInt
32
void canChange (int* x) {*x = 6;cout << *x << endl;
}void main ( ) {
int myInt = 17;int* ptr = &myInt;canChange (ptr);cout << myInt << endl;
}
Now with PointersNow with Pointers
0 1 2
3 4 5
6 7 8
-2 6 571
-2991 0 -33
41 412 -1
33
void canChange (int* x) {*x = 6;cout << *x << endl;
}void main ( ) {
int myInt = 17;int* ptr = &myInt;canChange (ptr);cout << myInt << endl;
}
Declare myIntDeclare myInt
0 1 2
3 4 5
6 7 8
-2 6 571
-2991 0 -33
41 17 -1
myInt
34
void canChange (int* x) {*x = 6;cout << *x << endl;
}void main ( ) {
int myInt = 17;int* ptr = &myInt;canChange (ptr);cout << myInt << endl;
}
Declare a Pointer to myIntDeclare a Pointer to myInt
0 1 2
3 4 5
6 7 8
7 6 571
-2991 0 -33
41 17 -1
myInt
ptr
35
void canChange (int* x) {*x = 6;cout << *x << endl;
}void main ( ) {
int myInt = 17;int* ptr = &myInt;canChange (ptr);cout << myInt << endl;
}
Pass a Copy of ptrPass a Copy of ptr
0 1 2
3 4 5
6 7 8
7 6 571
-2991 0 -33
41 17 -1
myInt
ptr
36
void canChange (int* x) {*x = 6;cout << *x << endl;
}void main ( ) {
int myInt = 17;int* ptr = &myInt;canChange (ptr);cout << myInt << endl;
}
Pass a Copy of ptrPass a Copy of ptr
0 1 2
3 4 5
6 7 8
7 6 7
-2991 0 -33
41 17 -1
myInt
ptr x
37
void canChange (int* x) {*x = 6;cout << *x << endl;
}void main ( ) {
int myInt = 17;int* ptr = &myInt;canChange (ptr);cout << myInt << endl;
}
Change Whatever x is Change Whatever x is Pointing tooPointing too
0 1 2
3 4 5
6 7 8
7 6 7
-2991 0 -33
41 17 -1
myInt
ptr x
38
void canChange (int* x) {*x = 6;cout << *x << endl;
}void main ( ) {
int myInt = 17;int* ptr = &myInt;canChange (ptr);cout << myInt << endl;
}
Change Whatever x is Change Whatever x is Pointing tooPointing too
0 1 2
3 4 5
6 7 8
7 6 7
-2991 0 -33
41 6 -1
myInt
ptr x
39
void canChange (int* x) {*x = 6;cout << *x << endl;
}void main ( ) {
int myInt = 17;int* ptr = &myInt;canChange (ptr);cout << myInt << endl;
}
Follow x and Print it Out (6)Follow x and Print it Out (6)
0 1 2
3 4 5
6 7 8
7 6 7
-2991 0 -33
41 6 -1
myInt
ptr x
40
void canChange (int* x) {*x = 6;cout << *x << endl;
}void main ( ) {
int myInt = 17;int* ptr = &myInt;canChange (ptr);cout << myInt << endl;
}
See the Change in main (6 See the Change in main (6 also)also)
0 1 2
3 4 5
6 7 8
7 6 7
-2991 0 -33
41 6 -1
myInt
ptr x
41
void canChange (int* x) {*x = 6;cout << *x << endl;
}void main ( ) {
int myInt = 17;int* ptr = &myInt;canChange (ptr);cout << myInt << endl;
}
Interesting NoteInteresting Note
At this point,these two statements print out the same thing!
cout << *ptr << endl;cout << myInt << endl
So do these!
cout << ptr << endl;cout << &myInt << endl;
WHY?
42
Allocating SpaceAllocating Space
• new vs. malloc ( );• delete vs. free( );
43
SummarySummary
• To understand pointers, you need to understand memory
• The & is the secret to it all!• Create and dereference with *• Passing a pointer to a function can
make changes to the main
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