CS31 Discussion 1E Spring 17: week 08 TA: Bo-Jhang Ho - - PowerPoint PPT Presentation

CS31 Discussion 1E

Spring 17’: week 08

TA: Bo-Jhang Ho

bojhang@cs.ucla.edu

Credit to former TA Chelsea Ju

Project 5 - Map cipher to crib

} Approach 1: For each pair of positions, check two letters

in cipher and crib are both identical or different

} For each pair of positions pos1 and pos2,

cipher[pos1] == cipher[pos2] should equal to crib[pos1] == crib[pos2] } Approach 2: Get the positions of the letter and compare

} For each position pos,

indexes1 = getAllPositions(cipher, pos, cipher[pos]) indexes2 = getAllPositions(crib, pos, crib[pos]) indexes1 should equal to indexes2

Project 5 - Map cipher to crib

} Approach 3: Generate the mapping

} We first have a mapping array char cipher2crib[128] = {‘\0’}; } Then whenever we attempt to map letterA in cipher to letterB

in crib, we first check whether it violates the previous setup:

} Is cipher2crib[letterA] != ‘\0’?

// implies letterA has been used

} Then cipher2crib[letterA] should equal to letterB

} If no violation happens,

} cipher2crib[letterA] = letterB;

Road map of CS31

Variable If / else Loops Pointer!! Array Function Class String

Cheat sheet

Data type

} int *a;

} a is a variable, whose data type is int * } a stores an address of some integer

“Use as verbs”

} & - address-of operator

} &b means I want to get the memory address of variable b

} * - dereference operator

} *c means I want to retrieve the value in address c

Memory model

Address Memory Contents

1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

1 byte

Memory model

Address Memory Contents

1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

1 byte = 8 bits

• ff
• n

A bit has 2 states

Memory model

Address Memory Contents

1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

1 byte = 8 bits

• ff
• n

A bit has 2 states A byte has 256 (=28) states

Memory model

Address Memory Contents

1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

1 byte = 8 bits

• ff
• n

A bit has 2 states A byte has 256 (=28) states A char takes 1 byte An int takes 4 bytes A double takes 8 bytes

Data type review

} Basic data types

Type Bytes Bits Value range char 1 8

• 128 to 127

short 2 16

• 32,768 to 32,767

int 4 32

• 2,147,483,648 to 2,147,483,647

long long 8 64

• 9 * 10^18 to 9 * 10^18

float 4 32

• 3.4 * 10^38 to 3.4 * 10^38

double 8 64

• 1.7 * 10^308 to 1.7 * 10^308

Memory model

Address Memory Contents

1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

1 byte = 8 bits

• ff
• n

A bit has 2 states A byte has 256 (=28) states My has 16GB ram

Memory model

Address Memory Contents

1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

1 byte = 8 bits

• ff
• n

A bit has 2 states A byte has 256 (=28) states My has 16GB ram = 16,000,000,000 bytes!

What’s going on in the memory

int main() { int a = 5; int b = 3; double c = 3.5; int d = b - a; return 0; } Address Memory Contents

1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

What’s going on in the memory

int main() { int a = 5; int b = 3; double c = 3.5; int d = b - a; return 0; } Address Memory Contents

1000

5

1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

a

What’s going on in the memory

int main() { int a = 5; int b = 3; double c = 3.5; int d = b - a; return 0; } Address Memory Contents

1000

5

1004

3

1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

a b

What’s going on in the memory

int main() { int a = 5; int b = 3; double c = 3.5; int d = b - a; return 0; } Address Memory Contents

1000

5

1004

3

1008

3.5

1016 1017 1018 1019

a b c

What’s going on in the memory

int main() { int a = 5; int b = 3; double c = 3.5; int d = b - a; return 0; } Address Memory Contents

1000

5

1004

3

1008

3.5

1016

2

a b c d

What’s going on in the memory

int main() { int a = 5; int *b = &a; a++; *b += 2; cout << a << endl; cout << *b << endl; return 0; } Address Memory Contents

1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

What’s going on in the memory

int main() { int a = 5; int *b = &a; a++; *b += 2; cout << a << endl; cout << *b << endl; return 0; } Address Memory Contents

1000

5

1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

a

What’s going on in the memory

int main() { int a = 5; int *b = &a; a++; *b += 2; cout << a << endl; cout << *b << endl; return 0; } Address Memory Contents

1000

5

1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

a

What’s going on in the memory

int main() { int a = 5; int *b; // declare a pointer var b = &a; // set address a++; *b += 2; cout << a << endl; cout << *b << endl; return 0; } Address Memory Contents

1000

5

1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

a

What’s going on in the memory

int main() { int a = 5; int *b; // declare a pointer var b = &a; // set address a++; *b += 2; cout << a << endl; cout << *b << endl; return 0; } Address Memory Contents

1000

5

1004

??

(address)

1012 1013 1014 1015 1016 1017 1018 1019

a b

What’s going on in the memory

int main() { int a = 5; int *b; // declare a pointer var b = &a; // set address a++; *b += 2; cout << a << endl; cout << *b << endl; return 0; } Address Memory Contents

1000

5

1004

1000

(address)

1012 1013 1014 1015 1016 1017 1018 1019

a b

What’s going on in the memory

int main() { int a = 5; int *b; // declare a pointer var b = &a; // set address a++; *b += 2; cout << a << endl; cout << *b << endl; return 0; } Address Memory Contents

1000

6

1004

1000

(address)

1012 1013 1014 1015 1016 1017 1018 1019

a b

5 à

What’s going on in the memory

int main() { int a = 5; int *b; // declare a pointer var b = &a; // set address a++; *b += 2; cout << a << endl; cout << *b << endl; return 0; } Address Memory Contents

1000

8

1004

1000

(address)

1012 1013 1014 1015 1016 1017 1018 1019

a b

6 à

What’s going on in the memory

int main() { int a = 5; int *b; // declare a pointer var b = &a; // set address a++; *b += 2; cout << a << endl; cout << *b << endl; return 0; } Address Memory Contents

1000

8

1004

1000

(address)

1012 1013 1014 1015 1016 1017 1018 1019

a b

} Output

} 8

What’s going on in the memory

int main() { int a = 5; int *b; // declare a pointer var b = &a; // set address a++; *b += 2; cout << a << endl; cout << *b << endl; return 0; } Address Memory Contents

1000

8

1004

1000

(address)

1012 1013 1014 1015 1016 1017 1018 1019

a b

} Output

} 8

8

Using uninitialized variables is always dangerous

int main() { int a; int b = a + 3; int *c; *c = 27; return 0; } Address Memory Contents

1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

Using uninitialized variables is always dangerous

int main() { int a; int b = a + 3; int *c; *c = 27; return 0; } Address Memory Contents

1000

??

1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

a

Using uninitialized variables is always dangerous

int main() { int a; int b = a + 3; int *c; *c = 27; return 0; } Address Memory Contents

1000

??

1004

??

1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019

a b

Using uninitialized variables is always dangerous

int main() { int a; int b = a + 3; int *c; *c = 27; return 0; } Address Memory Contents

1000

??

1004

??

1008

??

(address)

1016 1017 1018 1019

a b c

Using uninitialized variables is always dangerous

int main() { int a; int b = a + 3; int *c; *c = 27; return 0; } Address Memory Contents

1000

??

1004

??

1008

??

(address)

1016 1017 1018 1019

a b c

} May manipulate a piece of

memory not belonging to this program!!

Array v.s. Pointer

Array is a special case of pointer Pointer can be treated as an array

Array in memory

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = &b[1]; *c = 100; return 0; } Address Memory Contents

1000 1002 1004 1006 1008 1010 1012 1014 1016 1018 1020 1022 1024 1026 1028 1030 1032 1034 1036 1038

Array in memory

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = &b[1]; *c = 100; return 0; } Address Memory Contents

1000

3

1004 1006 1008 1010 1012 1014 1016 1018 1020 1022 1024 1026 1028 1030 1032 1034 1036 1038

a

Array in memory

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = &b[1]; *c = 100; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

30

1016

40

1020

50

1024 1026 1028 1030 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b

Array in memory

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = &b[1]; *c = 100; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

30

1016

40

1020

50

1024 1026 1028 1030 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b

Array in memory

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c; // declare a pointer c = &b[1]; // get address *c = 100; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

30

1016

40

1020

50

1024 1026 1028 1030 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b

Array in memory

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c; // declare a pointer c = &b[1]; // get address *c = 100; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

30

1016

40

1020

50

1024

??

(address) 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b c

Array in memory

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c; // declare a pointer c = &b[1]; // get address *c = 100; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

30

1016

40

1020

50

1024

1008

(address) 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b c

Array in memory

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c; // declare a pointer c = &b[1]; // get address *c = 100; return 0; } Address Memory Contents

1000

3

1004

10

1008

100

1012

30

1016

40

1020

50

1024

1008

(address) 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b c

20 à

Time and Delta analogy

Time Delta Time + = Delta Delta Delta + = Delta Time Time + = Time Time

Doesn’t make sense!

+ =

Time and Delta analogy

+ = + = + =

Doesn’t make sense!

+ =

Time and Delta analogy

Pointer Int Pointer + = Int Int Int + = Int Pointer Pointer + = Pointer Pointer

Doesn’t make sense!

+ =

Time and Delta analogy

Time Delta Time

• =

Delta Delta Delta

• =

Delta Time

• =

Time Time

• =

??

Doesn’t make sense!

Time and Delta analogy

Time Delta Time

• =

Delta Delta Delta

• =

Delta Time

• =

Time Time

• =

Delta

Doesn’t make sense!

Time and Delta analogy

= = = =

• Doesn’t

make sense!

Time and Delta analogy

Pointer Int Pointer

• =

Int Int Int

• =

Int Pointer

• =

Pointer Pointer

• =

Int

Doesn’t make sense!

Array in memory

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = &b[1]; c = c + 1; // Or, // c += 1; // c++; *c = 27; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

30

1016

40

1020

50

1024

1008

(address) 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b c

Array in memory

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = &b[1]; c = c + 1; // Or, // c += 1; // c++; *c = 27; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

30

1016

40

1020

50

1024

1012

(address) 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b c

1008à

} What? 1008 + 1 = 1012?

Array in memory

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = &b[1]; c = c + 1; // Or, // c += 1; // c++; *c = 27; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

30

1016

40

1020

50

1024

1012

(address) 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b c

1008 à

} Pointer arithmetic

} We should interpret as adding

the memory size of 1 integer

Array in memory

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = b; c += 2; *c = 27; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

30

1016

40

1020

50

1024

1004

(address) 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b c

} Array name is a pointer

} b can be treated as an int*

Array in memory

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = b; c += 2; *c = 27; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

30

1016

40

1020

50

1024

1012

(address) 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b c

} Array name is a pointer

} b can be treated as an int*

1004à

Array in memory

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = b; c += 2; *c = 27; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

27

1016

40

1020

50

1024

1012

(address) 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b c

} Array name is a pointer

} b can be treated as an int*

30 à

More about array

} Array name can be considered as start point

} Technically, it’s the base address

} The index can be considered as the offset

More about array

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = b; cout << b[-1] << endl; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

27

1016

40

1020

50

1024

1004

(address) 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b c

b[0] b[-1] b[1] b[2] b[3] b[4] b[5] b[6] b[7] b[8] b[9] b[-2] b[-3] b[-4]

Use pointer like an array

} If x is a pointer, you can treat x as an array

} Meaning, you can have something like x[3]

Treat pointer as an array

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = b + 2; cout << c[-2] << endl; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

27

1016

40

1020

50

1024

1012

(address) 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b c

Treat pointer as an array

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = b + 2; cout << c[-2] << endl; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

27

1016

40

1020

50

1024

1012

(address) 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b c

c[0]

Treat pointer as an array

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = b + 2; cout << c[-2] << endl; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

27

1016

40

1020

50

1024

1012

(address) 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b c

c[0] c[1] c[2] c[3] c[4] c[5] c[6] c[7]

Treat pointer as an array

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = b + 2; cout << c[-2] << endl; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

27

1016

40

1020

50

1024

1012

(address) 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b c

c[-2] c[-3] c[-1] c[0] c[1] c[2] c[3] c[4] c[5] c[6] c[7] c[-4] c[-6]

Treat pointer as an array

int main() { int a = 3; int b[5] = {10, 20, 30, 40, 50}; int *c = b + 2; cout << c[-2] << endl; return 0; } Address Memory Contents

1000

3

1004

10

1008

20

1012

27

1016

40

1020

50

1024

1012

(address) 1032 1034 1036 1038

a

b[0] b[1] b[2] b[3] b[4]

b c

c[-2] c[-3] c[-1] c[0] c[1] c[2] c[3] c[4] c[5] c[6] c[7] c[-4] c[-6]

} It is actually a valid

memory access

Summary

} Array name is a pointer } Pointer arithmetic

} Let’s say x is a pointer, n is an integer } x + n is a pointer (memory address) after n elements of x } x - n is a pointer before n elements of x

} Treat a pointer as an array

} Again, let’s say x is a pointer, n is an integer } x[n] means to access nth element counted from x } x[n] equivalent to *(x + n)

Remind a previous example 01 - Pass an array to a function

} From caller: } In the function:

• r
• r

int arr[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; foo(arr); void foo(int params[10]) { … } void foo(int params[]) { … } void foo(int *params) { … }

Remind a previous example 02 - Mapping of cipher and crib in project 5

} Approach 3: Generate the mapping

} We first have a mapping array char cipher2crib[128] = {‘\0’};

} Whenever we attempt to map letterA in cipher to letterB in crib: } cipher2crib[letterA] = letterB;

} But there are only 26 letters. Why do we want to have an array

with 128 elements?

} We use ascii as a key (a.k.a index) in the array } We don’t need to worry about the offset (i.e., x - ‘a’)

} Is there any solution to reduce the size to 26?

Remind a previous example 02 - Mapping of cipher and crib in project 5

bool check(const char cipher[], const char crib[]) { // assume cipher and crib have the same length char cipher2crib[128] = {'\0'}; for (int i = 0; cipher[i] != '\0'; i++) { char letterA = cipher[i]; char letterB = crib[i]; if (cipher2crib[letterA] != '\0’ && cipher2crib[letterA] != letterB) return false; cipher2crib[letterA] = letter } return true; }

Remind a previous example 02 - Mapping of cipher and crib in project 5

bool check(const char cipher[], const char crib[]) { // assume cipher and crib have the same length char trueArray[26] = {'\0'}; char *cipher2crib = trueArray – 'a'; for (int i = 0; cipher[i] != '\0'; i++) { char letterA = cipher[i]; char letterB = crib[i]; if (cipher2crib[letterA] != '\0’ && cipher2crib[letterA] != letterB) return false; cipher2crib[letterA] = letter } return true; }

26 elements trueArray cipher2crib

97 elements

Memory footprint

Question 01

} If we declare int *ptr; and int val;, the

following two usages are valid:

} ptr = &val; } *ptr = val;

} What is &ptr?

Question 01

} If we declare int *ptr; and int val;, the

following two usages are valid:

} ptr = &val; } *ptr = val;

} What is &ptr?

} &ptr means the address of ptr } The type of &ptr is int** (we call it double pointer) } For example, int** strongPtr = &ptr;

Double pointer

int main() { int a = 3; int *b = &a; int c[5] = {10, 20, 30, 40, 50}; int** d = &b; return 0; } Address Memory Contents

1000

3

1004 1006 1008 1010 1012 1014 1016 1018 1020 1022 1024 1026 1028 1030 1032 1034 1036 1038

a

Double pointer

int main() { int a = 3; int *b = &a; int c[5] = {10, 20, 30, 40, 50}; int** d = &b; return 0; } Address Memory Contents

1000

3

1004

1000

(address) 1012 1014 1016 1018 1020 1022 1024 1026 1028 1030 1032 1034 1036 1038

a b

Double pointer

int main() { int a = 3; int *b = &a; int c[5] = {10, 20, 30, 40, 50}; int** d = &b; return 0; } Address Memory Contents

1000

3

1004

1000

(address) 1012

10

1016

20

1020

30

1024

40

1028

50

1032 1034 1036 1038

a b

c[0] c[1] c[2] c[3] c[4]

c

Double pointer

int main() { int a = 3; int *b = &a; int c[5] = {10, 20, 30, 40, 50}; int** d = &b; return 0; } Address Memory Contents

1000

3

1004

1000

(address) 1012

10

1016

20

1020

30

1024

40

1028

50

1032

1004

(address)

a b

c[0] c[1] c[2] c[3] c[4]

c d

Question 02

} If we declare int *ptr; and int val;, the

following two usages are valid:

} ptr = &val; } *ptr = val;

} What is *val?

Question 02

} If we declare int *ptr; and int val;, the

following two usages are valid:

} ptr = &val; } *ptr = val;

} What is *val?

} It won’t compile } * operator (dereference) implies that what it stores is a

memory address

} Only pointer variables store memory address

Different levels of pointers

If we have

int val; int* ptr1; int** ptr2; int*** ptr3; int**** ptr4; int***** ptr5;

Then they have the following relations

ptr1 = &val; ptr2 = &ptr1; ptr3 = &ptr2; ptr4 = &ptr3; ptr5 = &ptr4; val = *ptr1; ptr1 = *ptr2; ptr2 = *ptr3; ptr3 = *ptr4; ptr4 = *ptr5;

Address of Deference

Question 03

} If we declare int *ptr;, can we hardcode an address

and assign to ptr?

} For example, ptr = 1234;

} No, it won’t compile

} For security issue } It doesn’t make sense that we can get an address beforehand } The same variable can reside in different parts of memory in

different executions

Question 03

int main() { int a; cout << &a << endl; return 0; }

Question 04

} If we declare int *ptr; and double val;, is the

following code valid?

} ptr = &val;

Question 04

} If we declare int *ptr; and double val;, is the

following code valid?

} ptr = &val;

} No, it won’t compile

} Pointers are type-aware } We can cast the type: ptr = (int*) &val; } However, that means we use the way we interpret integer to

intepret a piece of memory which stores a double

Array in memory

int main() { double a = 3.5; int *b = (int*) &a; cout << *b << endl; return 0; } Address Memory Contents

1000

3.5

1008

1000

(address) 1016 1018 1020 1022 1024 1026 1028 1030 1032 1034 1036 1038

a b

Array in memory

int main() { double a = 3.5; int *b = (int*) &a; cout << *b << endl; return 0; } Address Memory Contents

1000

3.5

1008

1000

(address) 1016 1018 1020 1022 1024 1026 1028 1030 1032 1034 1036 1038

a b

Question 05

} If we declare int *ptrI; and double *ptrD;,

can we have the following assignment?

} ptrI = ptrD;

} No, it won’t compile

} Pointer type doesn’t match } Though both store memory addresses, how they interpret the

memory content are different

} We can cast the type: ptrI = (int*) ptrD;

Checkpoint – Looping over the array

double* findFirstNegativePtr(double a[], int n) { for (double* p = a; p < a + n; p++) { if (*p < 0) return p; } return nullptr; }

} Return a pointer

int findFirstNegativeIdx(double a[], int n) { for (int i = 0; i < n; i++) { if (a[i] < 0) return i; } return -1; }

} Return an index

Project 6

} Problem 1b probably is the most tricky question.

Road map of CS31

Variable If / else Loops Pointer!! Array Function Class String

Class

} Define a data structure } A data structure groups different “variables” together } For example, when we describe a 2d coordinate, naturally

we use 2 numbers to represent it

} We can also say we declare a new data type

Example

// create a new data type class Point { public: double x; double y; }; // how we use it int main() { Point p; p.x = 1.1; p.y = 2.2; Point r = p; r.y = 3.3; cout << "Point 1: " << p.x << " " << p.y << endl; cout << "Point 2: " << r.x << " " << r.y << endl; return 0; }