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REACH CECS 130 Final Test Review

REACH CECS 130 Final Test Review

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REACH CECS 130 Final Test Review

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  1. REACH CECS 130 Final Test Review

  2. Data • How a computer stores data in its internal memory • RAM (Random-Access Memory) - temporary • ROM (Read-Only Memory) – non volatile • Store data in bytes • How you store data temporarily • Create variables based on fundamental types (bool, char, int, float) • constants: #define CONSTNAME value • sizeof()

  3. Variable Types

  4. Control Statements – Boolean Operators • What do each of the following evaluate to? 1. long elves = 8; int dwarves = 8; if(elves==dwarves) //true or false? if(elves!=0) //true or false? 2. int elves = 4; int dwarves = 5; if(dwarves > (2/3)) //true or false? 3. if(0 < x < 99) //true or false? 4. if(0<= (0<1))//true or false?

  5. Control Statements – Boolean Operators -Answers • What do each of the following evaluate to? 1. long elves = 8; int dwarves = 8; if(elves==dwarves) //true if(elves!=0) //true 2. int elves = 4; int dwarves = 5; if(dwarves > (2/3)) //true 3. if(0 < x < 99) //true …TRUE (1) and FALSE (0) < 99 4. if(0<= (0<1))//true

  6. If statements • if(condition) statement; else if (condition) statement; • condition ? expr1 : expr2 • ex. z = ( x > y ) ? y : x ; • Can we do next statement? (x>y) ? cout << “x is greater than y.” : cout << “x isn’t greater than y.”;

  7. Switch-case statements switch(expression){ case expr1: statement; break; case expr2: statement; break; case expr3: statement; break; default: statements break; }

  8. Loops • while (condition) { statements; You need to maintain a state to break loop. } • do { statements; You need to maintain a state to break loop. } while(condition);

  9. Loops • for (initialization; condition; expression) { statements; } • Incrementing: Prefix and Postfix int x = 5; int y = 6; int z = y++ //z=6, y=7 postfix operator int z = ++x //z=6, x=6 prefix operator

  10. Branching statements

  11. FOR LOOP TEST • Can you write a program that prints out the following? 0 1 2 3 4 5 6 7 8 9

  12. Answer for ( int count = 0; count < 10; count ++) { cout <<count<<“”; }

  13. Other Challenge questions for Loops • Write a conditional statement that will assign x/y to x if y doesn’t equal 0. • Write a while loop that calculates the summation of positive integers from 1 to some number n. • Write a conditional statement that assigns x*y if x is even; otherwise , if x is odd and y doesn’t equal 0, assign x to x/y; if neither of the preceding cases is true, output to the screen that y is equal to 0.

  14. How to declare and implement functions • Function declaration • Function definition • Function call

  15. #include <iostream> using namespace std; int add(int, int); int main(void) { int number1, number2; cout << “Enter the first value to be summed:”; cin >> number1; cout << “\nEnter the second:”; cin >> number2; cout << “\n The sum is: “ << add (number1, number2) <<endl; } int add(int a, int b){return a+b;}

  16. Functions Challenge • Write a function, called multiply that multiplies two numbers and returns the result

  17. Union Output: 9 3.1416 #include <stdio.h> #include<stdlib.h> union NumericType { int iValue; long lValue; double dValue; }; int main() { union NumericType Values; // iValue = 10 Values.iValue=9; printf("%d\n", Values.iValue); Values.dValue = 3.1416; printf("%f\n", Values.dValue); system("pause"); }

  18. Sizeof()

  19. Malloc() Please enter how long your name is: 21 Please enter your name: Nawaf Hello Nawaf Please enter how long your name is: -7 Failed allocation memory

  20. free()

  21. Calloc and Realloc() int *n; int * n1; n=( int * ) calloc(5, sizeof(int)); // Reserves a block of memory for 5 integers //Decide you need to reallocate more memory later in the program n1= (int *) realloc(n, 10 * sizeof(int));//allocate 10 integers instead of 5 if (n1!=NULL) { n=n1; } else printf("Out of memory!"); realloc() returns null if unable to complete or a pointer to the newly reallocated memory.

  22. C - Files Do you know the syntax for each of these, used to read and write to data files? • Pointers: think of it as the memory address of the file • fopen() • fclose() • fscanf() • fprintf()

  23. fopen(“file name”, “Mode”) • fopen() returns a FILE pointer back to the pRead variable • #include <cstdio> • Main() • { • FILE *pRead; • pRead = fopen(“file1.dat”, “r”); • if(pRead == NULL) • printf(“\nFile cannot be opened\n”); • else • printf(“\nFile opened for reading\n”); • fclose(pRead); • }

  24. What does this code do? int main () { FILE * pFile; char c; pFile=fopen("alphabet.txt","wt"); for (c = 'A' ; c <= 'Z' ; c++) { putc (c , pFile);//works like fprintf } fclose (pFile); return 0; }

  25. Common Text File Modes

  26. fclose(file pointer) • Pretty basic. • Always close files when you use fopen.

  27. fscanf(FILE pointer, “data type”, variable in which to store the value) • Reads a single field from a data file • “%s” will read a series of characters until a white space is found • can do fscanf(pRead, “%s%s”, name, hobby);

  28. #include <stdio.h> • Main() • { • FILE *pRead; • char name[10]; • pRead = fopen(“names.dat”, “r”); • if( pRead == NULL ) • printf( “\nFile cannot be opened\n”); • else • printf(“\nContents of names.dat\n”); • fscanf( pRead, “%s”, name ); • while( !feof(pRead) ) { // While end of file not reached • printf( “%s\n”, name ); // output content of name • fscanf( pRead, “%s”, name ); // scan from file next string • } • fclose(pRead); • }

  29. Quiz Kelly 11/12/86 6 Louisville Allen 04/05/77 49 Atlanta Chelsea 03/30/90 12 Charleston Can you write a program that prints out the contents of this information.dat file?

  30. #include <stdio.h> • Main() • { • FILE *pRead; • char name[10]; • char birthdate[9]; • float number; • char hometown[20]; • pRead = fopen(“information.dat”, “r”); • if( pRead == NULL ) • printf( “\nFile cannot be opened\n”); • else • fscanf( pRead, “%s%s%f%s”, name, birthdate, &number, hometown ); • while( !feof(pRead) ) { • printf( “%s \t %s \t %f \t %s\n”, name, birthdate, number, hometown ); • fscanf( pRead, “%s%s%f%s”, name, birthdate, &number, hometown ); • } • fclose(pRead); • }

  31. fprintf(FILE pointer, “list of data types”,list of values or variables) • The fprintf() function sends information (the arguments) according to the specified format to the file indicated by stream. fprintf() works just like printf() as far as the format goes.

  32. #include <stdio.h> Main() { FILE *pWrite; char fName[20]; char lName [20]; float gpa; pWrite = fopen(“students.dat”,”w”); if( pWrite == NULL ) printf(“\nFile not opened\n”); else printf(“\nEnter first name, last name, and GPA ”); printf(“separated by spaces:”); scanf(“%s%s%f”, fName, lName, &gpa); fprintf(pWrite, “%s \t %s \t % .2f \n”, fName, lName, gpa); fclose(pWrite); }

  33. Quiz • Can you write a program that asks the user for their • Name • Phone Number • Bank account balance And then prints this information to a data file called accounts.dat ?

  34. C++ Input/Output • Summary • Include #include <iostream> directive at beginning of program • Use cin to take data from user • Use cout to display data on screen • Display multiple strings and integers in the same cout statement by separating items with <<

  35. C++ Input/Output Example #include <iostream> #include<string> using namespace std; string name = “”; int main(void) { cout<<“What is your name?”; cin>>name; cout<<endl<<“Hello”<<name.c_str(); return 0; }

  36. Can you predict the printout? #include <iostream> using namespace std; int x = 25; string str2 = “This is a test”; int main( void ) { cout<<“Test”<<1<<2<<“3”; cout<<25 %7<<endl<<str2.c_str(); return 0; }

  37. Answer Test 1234 This is a test

  38. OOP • Declare classes • Create objects • 3 MAIN PRINCIPLES OF OOP • Data abstraction – hiding data members and implementation of a class behind an interface so that the user of the class corrupt that data • Encapsulation – each class represents a specific thing or concept. Multiple classes combine to produce the whole • Polymorphism-objects can be used in more than one program

  39. Classes • Classes are general models from which you can create objects • Classes have data members either data types or methods • Classes should contain a constructor method and a destructor method • See handout for example of a program that utilizes a class

  40. Declaring Classes class ClassName { memberList }; memberList can be either data member declarations or method declarations

  41. Class Declaration Example Class Bow { //data member declarations string color; bool drawn; intnumOfArrows; Bow(string aColor); //constructor ~Bow(); //destructor //methods void draw(); int fire(); };

  42. Creating Methods Return_type ClassName::methodName(argumentList) { methodImplementation }

  43. Methods Creation Example //draws the bow Void Bow::draw() { drawn = true; cout<< “The “<<color<<“bow has been drawn.”<<endl; }

  44. Advanced Data Types • Arrays • Pointers

  45. How to create arrays • data_typearray_name [number-of-elements]; • Two Dimensional Array array_typearray_name [number_of_ROWS][number_of_COLUMNS];

  46. How to create pointers • type* pointer_name; • ex. int my_int; int* my_int_pointer = &my_int; Assigns the address of my_int to the pointer

  47. Useful string functions • Copying strings from one to another • char* strcpy(char* p, const char* q); • char s[6]; strcpy(s, “Hello”); • To combine strings • char* strcat(char* p, const char* q); • char s[12] = “Hello” strcat(s, “World”);

  48. Useful string functions • To copy n characters from q to the of p. • char* strncpy(char* p, const char* q, int n); • char s [7] = “Say “; char t[] = “Hi”; strncpy (s, t, 2) Output: Hiy

  49. Challenge question • Can you write a program using C++ that uses a FOR loop to initialize a 2D array that looks like the following {0,5,10,15}{0,2,4,6}

  50. Answer #include<iostream> using namespace std; int main(){ int array[2][4], , row, column; for(row=0;row<2;row++) for(column=0;column<4;column++){ if(row==0) array[row][column]=column*5; else if(row==1) array[row][column]=column*2; } for(row=0; row<2; row++){ for(column =0; column <4; column ++) cout<<array[row][column]<<" "; cout<<endl; } system("pause"); return 0; }