C++课件(精华版)

1、2、c+ programming,a simple c+ program types, variables, expressions,how to write a program,find the whole steps in the real model use some graph or nature language to describe realize with computer language,workflow,pseudo code,a first idea: int main() variables while (condition) analyze the expressi

2、on evaluate the expression print the result,4,main function,int main(void) return 0;,int x=5; int y=8; int z=x+y; coutx“+”y“=“zendl,5,2.1.cpp,include using namespace std; /* * a simple program for demonstrating the basics of a c+ project. * * it does a good job of demonstrating c+ fundamentals, but

3、a * terrible job with the poetry. */ int main() cout dont you just feel like a louse; cout endl; cout to learn that your new theorem was proved by gauss?; cout endl; return 0;,6,cout,cout,7,c+ tokens,a token is the smallest element of a c+ program that is meaningful to the compiler. kinds of tokens:

4、 identifiers, keywords, literals, operators, punctuators, and other separators. tokens are usually separated by white space. white space can be one or more blanks, horizontal or vertical tabs, new lines, form feeds or comments,8,c+ keywords,auto const double float int short struct unsigned unsigned

5、break break continue elsefor long switch void case sizeof typedef char do if return static union while ,etc,9,commenting,* *name of program *information of author *function of program * */ /a sample int main() / /*this is in the comment this is also in the comment */ .,10,constants,1,2,3 1.2, 4.50 “

6、name”, “your_phonenumber” ture,false 0 x12 1,a,$,xhh,ddd #define pi 3.141592 #define price 100 const int pi= 3.141592,11,2.2.cpp,include using namespace std; int main() int x; int y; x = 3; y = 4; cout x+y endl; return 0;,12,variables types,built-in types boolean type bool 1byte character types char

7、 1byte integer types int 2-4bytes(2) -3276832767 short (2) true, false character literals: char c; a, x, 4, n, $ integer literals: int x; 0, 1, 123, -6, 0 x34, 0 xa3 floating point literals: double d; float f; 1.2, 13.345, .3, -0.54, 1.2e3, . 3f, .3f string literals: string s; asdf, howdy, all yall,

8、14,variables names,choose meaningful names confuse mtbf, tla, myw, nbv short names can be meaningful x is a local variable i is a loop index dont use long names ok: partial_sum, element_count, staple_partition too long: the_number_of_elementsremaining_free_slots_in_the_symbol_table,15,not variables

9、names,a name in a c+ program starts with a letter, contains letters, digits, and underscores (only) x, number_of_elements, fourier_transform, z2 not names: 12x, time$to$market, main line not start names with underscores: _foo not use keywords int if while,16,declaration and initialization,int a = 7;

10、 int b = 9; char c = a; double x = 1.2; string s1 = hello, world; string s2 = 1.2,17,9,a,1.2,13 hello, world,4 | 1.2,b,c,x,s1,s2,7,a,constant variables,const int i=5; i=6; /error,18,think about,int a,b,c=2; int x,y,z,10; int m=2; int n=3; long int sum=0,add; long hello; char a=m; char b,c,d; char m=

11、65,n=a+1; float a,b,ccc=3.1415; float sum=0.0; double f1, f2=1.414e12,19,assignment and increment,int a = 7; a = 9; a = a+a; a += 2; +a,20,7,9,18,20,21,a,think about,int a=10, b; b=a; float x; int k=300; x=k; float x=3.14; int n; n=x+6; float x=3.14; int n; n=3; cout x+n; 3.0/9 or (float)3/9,2.3.cpp

12、 a program to illustrate integer overflow,include using namespace std; /* * a program to illustrate what happens when large integers * are multiplied together. */ int main() int million = 1000000; int trillion = million * million; cout according to this computer, million squared is trillion . endl;,

13、22,a type-safety violation(“implicit narrowing”,int main() int a = 20000; char c = a; int b = c; if (a != b) cout oops!: a != b n; else cout wow! we have large charactersn;,23,c+ character set,0 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z a b c d e f g h i j k l m n o p q r

14、 s t u v w x y z _ $ # ( ) % : ; . ? * + / char let = 97; cycle through the ascii table with math! char bee = a + 1,26,operators, + -, - * / ,27,arithmetic assignment operators,28,think about: int a=12; a += a -= a * a ; cout a,29,expressions,boolean type: bool (true and false) equality operators:=

15、= (equal), != (not equal) logical operators: / a program to investigate the behavior of the mod (%) operation. int main() int a, b; cout ; cin a; cout ; cin b; cout a % b = a%b endl; return 0;,30,think about,17/3=5? 5/9=0? int n; n%2+(n+1)%2=? n=2; n+; n=n+1 n=2: n+; n-; +n;-n; r=2; m=-n; p=r+; m=10

16、,n=100; p=(n+n,n*n,n-2); p=n+n,n*n,n-2,31,think about,int i=2; (i+)+(i+)+(i+) ; couti; i=2; (-i)+(-i) ; couti; i=2; i=(i+i+i) ; couti; i=2; i=(i-i) ; couti,2.6.cpp string input,include #include using namespace std; int main() cout first second; string name = first + + second; cout hello, name n;,33,

17、think about: the output,int x,y; x = 10; y = x+; cout y endl; int a,b; a = 10; b = +a; cout b endl,34,boolean expressions,a b a a=! k,37,38,statements,a = b; double d2 = 2.5; if (x = 2) y = 4; while (true) cout“hello”; for(int i=0;i8;i+) cout i; int average = (length+width)/2; return x,floor,int flo

18、or(float x) return (int) x;,39,ceiling,int ceiling(float x) if (x (int) x 0.0) return (int)(x + 1.0); else return (int) x;,40,to be continued,3、c+ programming,control statements,conditions1,if (condition) /do this,void main() int x; cin x; if (x0)x=-x; cout x; coutsqrt(x);,bool leapyear(int y,*any y

19、ear divisible by 4 except centenary years not divisible by 400*/ bool leapyear(int y) / any year divisible by 4 except centenary years not divisible by 400 if (y%4) return false; if (y%100=0,conditions2,if (condition) /do this else /do that,if (x0) cout-x; else coutx,void main() int a=5,b=8; if (ab)

20、 cout a; else cout b;,example,void main() char a; couta; if(a=a,conditions3,if (condition) /do this else if (condition) /do that else /do other thing,void main() int a=3,b=17,c=5; if (ab) if (bc) coutc) coutc) coutc) coutbca; else coutcba;,1 x0 y=f(x)= 0 x=0 -1 x0) k=1; else if (x=0) k=0; else k=-1; coutx k;,example,xy)? x:y,get the max one of a and b int max(int a, int b) int m = a b ? a : b ; return m;,if (condition) /do this else