struct Pila { private: int size; int defaultGrowthSize; int marker; int * contenuto;
CS31 Discussion 1E - CS | Computer...
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CS31 Discussion 1E Spring 17’: week 08 TA: Bo-Jhang Ho [email protected] Credit to former TA Chelsea Ju
Transcript of CS31 Discussion 1E - CS | Computer...
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
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1 byte
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1 byte = 8 bits
off on
A bit has 2 states
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1 byte = 8 bits
off on
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Memory modelAddress Memory Contents
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1 byte = 8 bits
off on
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 typesType 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 modelAddress Memory Contents
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1 byte = 8 bits
off on
A bit has 2 states
A byte has 256 (=28) states
My has 16GB ram
Memory modelAddress Memory Contents
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1 byte = 8 bits
off on
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;}
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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;}
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int main() {int a = 5;int b = 3;double c = 3.5;int d = b - a;
return 0;}
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int main() {int a = 5;int b = 3;double c = 3.5;int d = b - a;
return 0;}
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int main() {int a = 5;int b = 3;double c = 3.5;int d = b - a;
return 0;}
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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;}
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int main() {int a = 5;int *b = &a;
a++;*b += 2;
cout << a << endl;cout << *b << endl;
return 0;}
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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;}
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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;}
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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;}
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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;}
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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;}
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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;}
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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;}
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} 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;}
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} Output} 8
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Using uninitialized variablesis always dangerous
int main() {int a;int b = a + 3;int *c;
*c = 27;
return 0;}
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Using uninitialized variablesis always dangerous
int main() {int a;int b = a + 3;int *c;
*c = 27;
return 0;}
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Using uninitialized variablesis always dangerous
int main() {int a;int b = a + 3;int *c;
*c = 27;
return 0;}
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Using uninitialized variablesis always dangerous
int main() {int a;int b = a + 3;int *c;
*c = 27;
return 0;}
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Using uninitialized variablesis always dangerous
int main() {int a;int b = a + 3;int *c;
*c = 27;
return 0;}
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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;}
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Array in memory
int main() {int a = 3;int b[5] = {10, 20, 30, 40, 50};int *c = &b[1];
*c = 100;
return 0;}
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int main() {int a = 3;int b[5] = {10, 20, 30, 40, 50};int *c = &b[1];
*c = 100;
return 0;}
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Array in memory
int main() {int a = 3;int b[5] = {10, 20, 30, 40, 50};int *c = &b[1];
*c = 100;
return 0;}
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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;}
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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;}
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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;}
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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;}
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20 à
Time and Delta analogy
Time Delta Time + =
Delta Delta Delta+ =
Delta Time Time+ =
Time Time Doesn’tmake sense!+ =
Time and Delta analogy
+ =
+ =
+ =
Doesn’tmake sense!+ =
Time and Delta analogy
Pointer Int Pointer+ =
Int Int Int+ =
Int Pointer Pointer+ =
Pointer Pointer Doesn’tmake sense!+ =
Time and Delta analogy
Time Delta Time - =
Delta Delta Delta- =
Delta Time- =
Time Time- = ??
Doesn’tmake sense!
Time and Delta analogy
Time Delta Time - =
Delta Delta Delta- =
Delta Time- =
Time Time- = Delta
Doesn’tmake sense!
Time and Delta analogy
=
=
=
=
-
-
-
-
Doesn’tmake sense!
Time and Delta analogy
Pointer Int Pointer- =
Int Int Int- =
Int Pointer- =
Pointer Pointer- = Int
Doesn’tmake 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;}
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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;}
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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;}
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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;}
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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;}
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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;}
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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;}
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b[0]
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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;}
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b[2]
b[3]
b[4]
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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 Contents1000
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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 Contents1000
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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 Contents1000
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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 Contents1000
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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:
or
or
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
trueArraycipher2crib
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 Contents1000
31004
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 Contents1000
31004
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 Contents1000
31004
1000(address)
101210
101620
102030
102440
102850
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 Contents1000
31004
1000(address)
101210
101620
102030
102440
102850
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 pointersIf 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 Contents1000
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 Contents1000
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 typeclass Point {
public:double x;double y;
};
// how we use itint 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;}
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