title: C
#+STARTUP: content
Program
Decimal to Binary Octal Hexdecimal char
#include <stdio.h>
int main( )
{
int a = 80;
printf("the input a is %d\n",a);
printf("converse to 8 base is %#o\n", a);
printf("converse to 16 base is %#X\n", a);
printf("converse to char is %c\n", a);
return 0;
}
Primzahl filter
#include<stdio.h>
#include"math.h"
int main( )
{
int i;
int j;
for (i = 2; i < 10; i++) {
int temp = (int)sqrt(i);
for (j = 1; j <= 1000; j++) {
if (j != 1 && i%j == 0 ) {
break;
}else {
if (j == temp) {
printf("%d \n",i);
break;
}
}
}
}
return 0;
}
itoa
atoi, 如有非数字字符，就为0 itoa , 非ASCII标准函数库，看collection.
#include <stdio.h>
#include <stdlib.h>
char* itoa(int val, int base){
static char buf[32] = {0};
int i = 30;
for(; val && i ; --i, val /= base)
buf[i] = "0123456789abcdef"[val % base];
return &buf[i+1];
}
int main(int argc, char *argv[])
{
int a = 100;
char* str = itoa(a, 8);
printf("%s \n",str);
return 0;
}
each time only reserive 3 agruments
- Each time ask default 3 agreements, less is ok, the more will be ignored.
- scanf("%s", a); can't deal with more agreements. the more will be read next time
- scanf("%[^
\n{=latex}]", a); can't deal with less agreements,
because without \n{=latex}, I don't know where to stop
- scanf("c%",a); can use '
\n{=latex}' and ' ' to determine the stop and number of agreements.
#include <stdio.h>
#include <string.h>
int main(int argc, char *argv[])
{
while(1){
char a[3][10] = {"", "", ""};
char b[100];
scanf("%[^\n]", b);
printf("b is %s\n", b);
int j = 0;
int i = 0;
do{
int k = 0;
do{
a[j][k] = b[i];
i++;
k++;
}while(b[i] != ' ');
printf("a[%d] iooooos %s\n", j, a[j]);
j++;
i++;
} while (j<3);
scanf("%*[^\n]"); scanf("%*c");
}
// return 0;
}
SortingGiveNumbersWithExchangeSortedDirection
milisecond time
#include <sys/time.h>
#include <stdlib.h>
#include <stdio.h>
#include <unistd.h>
#define N 1000000
int current_timestamp() {
struct timeval te;
gettimeofday(&te, NULL); // get current time
int milliseconds = te.tv_sec*1000 + te.tv_usec/1000; // calculate milliseconds
return milliseconds;
}
int main(int argc, char *argv[])
{
float arr[N];
float summation;
int start, stop;
for (int i = 0; i < N; ++i) {
arr[i] = rand() % 100 / (float)100 ;
}
start = current_timestamp();
for (int i = 0; i < N; ++i) {
summation += arr[i];
}
stop = current_timestamp();
printf("Summation is %f, using time %d ", summation, stop-start);
return 0;
}
gcc task1_q1.c -o task1_q1
./task1_q1
Summation is 494979.375000, using time 4
Point&array
pointer assignment
int *c
only exits *c, if assige 1 to c, segment default
int *c;
printf("format string is %d, %d, %d\n" , *c, c, &c);
*c = 101;
printf("format string is %d, %d, %d\n" , *c, c, &c);#+end_src
#+RESULTS:
: format string is 1, -1754099728, -1754099984
: format string is 101, -1754099728, -1754099984
segment default for this
#+begin_src C :results output
int *c;
printf("format string is %d, %d, %d\n" , *c, c, &c);
c = 101;
printf("format string is %d, %d, %d\n" , *c, c, &c);
int * c = 100
only exit c, if access to *c, segment default
int *cc = 100;
printf("format string is %d, %d\n" , cc, &cc);
cc = 101;
printf("format string is %d, %d\n" , cc, &cc);
assign a address, all exits now
point to pass
void output(int *t){
printf("%d\n", *t );
(*t)++;
}
int c = 1;
printf("c : %d\n",c );
for (int i = 0; i< 10; i++){
output(&c);
}
printf("c : %d\n",c );
void output(int t[2]){
printf("%d\n", t[1] );
(t[1])++;
t[0] = 100;
}
int c[2] = {1,2};
printf("c[1] : %d\n",c[1] );
for (int i = 0; i< 10; i++){
output(c);
}
printf("c[0] : %d\n",c[0] );
printf("c[1] : %d\n",c[1] );
reference to pass not to that
void output(int &t){
printf("%d\n", t );
t++;
}
int a = 10;
printf("a : %d\n",a );
for (int i = 0; i< 10; i++){
output(a);
}
printf("a : %d\n",a );
char *ptr;
all exist
char *ptr ;
printf("format string %d is %s at %d \n" , *ptr, ptr, &ptr);
ptr = "lko";
printf("format string is %s \n" , ptr);
all exist
char *ptrr = "ok";
printf("format string %d is %s at %d \n" , *ptrr, ptrr, &ptrr);
ptrr = "lko";
printf("format string is %s \n" , ptrr);
pointer and array description
*p = a[n] p a
the first Value *p / p[0] *a / a[0] the n-th Value *(p+n)/ p[n] *a+n/ a[n] the first Address p / a / &a[0] the n-th Address p+n / &a[n]
在传递过程中， 数组的传递可以用指针来接受， 指针的传递可以用数组来接受，但必须是指针类型的数组
#include <iostream>
int main(){
int a[]={1,2,3,4,5};
int *p = a;
printf("Print this hallo!\n");
printf("for value\n");
printf("%d\n",*p);
printf("%d\n",*a);
printf("%d\n",a[0]);
printf("%d\n",p[0]);
printf("nihao :%d\n",*(p+1));
printf("%d\n",*a+1);
printf("%d\n",a[1]+1);
printf("%d\n",p[1]+1);
printf("for address:\n");
printf("%d\n",p);
printf("%d\n",a);
printf("%d\n",&a[0]);
printf("%d\n",p+2);
printf("%d\n",&a[2]);
printf("%c\n","0123456789abcdef"[3]);
char *list = "0123456789abcdef";
printf("%s\n",&list[10]);
return 0;
}
2orderPoint assignment to *
2 order Point assignment to (*a, a, &a)
#include <stdio.h>
void Point2order(int **a){
printf("**a : %d\n",**a );
printf("*a : %d\n",*a );
printf(" a : %d\n", a );
printf("&a : %d\n",&a );
printf("\n");
int *z = *a;
printf(" int *z = *a : if z = *a\n");
printf("z : %d\n", z );
printf("*a : %d\n", *a );
printf("if *z = **a\n");
printf("*z :%d\n", *z );
printf("**a:%d\n", **a );
printf("\n");
int *y = a;
printf(" int *y = a : if y = a\n");
printf("y : %d\n", y );
printf("a : %d\n", a );
printf("if *y = *a\n");
printf("*y : %d\n", *y );
printf("*a : %d\n", *a );
printf("\n");
int *x = &a;
printf(" int *x = &a : if x = &a\n");
printf("x : %d\n", x );
printf("&a : %d\n", &a );
printf("if *x = *(&a)\n");
printf("*x : %d\n", *x );
printf("*&a: %d\n", *(&a) );
}
int main(int argc, char *argv[])
{
int aa[] = {10,20,30};
int *a = aa;
Point2order(&a);
return 0;
}
**a : 10
*a : 1886852988
a : 1886852976
&a : 1886852904
int *z = *a : if z = *a
z : 1886852988
*a : 1886852988
if *z = **a
*z :10
**a:10
int *y = a : if y = a
y : 1886852976
a : 1886852976
if *y = *a
*y : 1886852988
*a : 1886852988
int *x = &a : if x = &a
x : 1886852904
&a : 1886852904
if *x = *(&a)
*x : 1886852976
*&a: 1886852976
Important
#include <stdio.h>
int main(){
int a[3][4]={0,1,2,3,4,5,6,7,8,9,10,11};
int(*p)[4];
int i,j;
p=a;
for(i=0; i<3; i++){
for(j=0; j<4; j++) printf("%2d ",*(*(p+i)+j));
printf("\n");
}
printf("\n");
int *m[4] = {a[0],a[1],a[2],a[3]};
for(i=0; i<3; i++){
for(j=0; j<4; j++) printf("%2d ",*(*(m+i)+j));
printf("\n");
}
return 0;
}
0 1 2 3
4 5 6 7
8 9 10 11
0 1 2 3
4 5 6 7
8 9 10 11
* assignment to 2orderPoint
char / int 数组/指针数组 to 二级指针 [] > * > + *(p+1)== p[1] and *p+1 == (*p)+1
#include <stdio.h>
void funaa(char **p){
printf("address mani after 2\n");
printf("funaa : %c\n", *p[0] );
}
void funbb(int **p){
printf("address mani after 2\n");
printf("funbb : %d\n", *p[0] );
}
void funa(char **p){
printf("value mani: \n");
printf("funa: the first %c\n", **p );
printf("funa: the third %c\n", *( (*p+1) +1) );
char *pa = *p+2;
funaa(&pa);
}
void funb(int **p){
printf("\n");
printf("value mani: \n");
printf("funb: the first %d\n", **p );
printf("funa: the third %d\n", *( (*p+1) +1) );
int *pa = *p +2;
funbb(&pa);
}
void funChar2OrderPoint(char ** p){
printf("\n");
printf("point arrar[0][0]:%c\n", *( *(p+0) +0) );
printf("point arrar[0][1]:%c\n", p[0][1] );
printf("point arrar[0][2]:%c\n", *( *(p+0) +2) );
printf("point arrar[1][0]:%c\n", *( *(p+1) +0) );
printf("point arrar[1][1]:%c\n", *( *(p+1) +1) );
printf("point arrar[1][2]:%c\n", *( *(p+1) +2) );
printf("point arrar[2][0]:%c\n", *( *(p+2) +0) );
printf("point arrar[2][1]:%c\n", *( *(p+2) +1) );
printf("point arrar[2][2]:%c\n", *( *(p+2) +2) );
}
void funInt2OrderPoint(int ** p){
printf("\n");
printf("point arrar[0][0]:%d\n", *( *(p+0) +0) );
printf("point arrar[0][1]:%d\n", *( *(p+0) +1) );
printf("point arrar[0][2]:%d\n", p[0][2] );
printf("point arrar[1][0]:%d\n", *( *(p+1) +0) );
printf("point arrar[1][1]:%d\n", *( *(p+1) +1) );
printf("point arrar[1][2]:%d\n", *( *(p+1) +2) );
printf("point arrar[2][0]:%d\n", *( *(p+2) +0) );
printf("point arrar[2][1]:%d\n", *( *(p+2) +1) );
printf("point arrar[2][2]:%d\n", *( *(p+2) +2) );
}
int main(int argc, char *argv[])
{
/* char 一维数组转二级指针 */
char a[] ="1234567";
char *pa = a;
funa(&pa);
/* int 一维数组转二级指针 */
int b[] = {1,2,3,4,5,6,7};
int *pb = b;
funb(&pb);
/* char 指针数组转二级指针 */
char a1[] ="123";
char a2[] = "456";
char a3[] = "789";
char *chara [] = {"123", "456", "789"};
char **poa = chara;
funChar2OrderPoint(poa);
printf("\n");
printf("%s\n", *chara+2 );
printf("%s\n", *chara+1 );
printf("%s\n", *chara );
printf("\n");
printf("%s\n", chara[0] );
printf("%s\n", chara[1] );
printf("%s\n", chara[2] );
printf("\n");
printf("%c\n", chara[1][2] );
printf("%c\n",*( chara[1] +2) );
printf("%c\n",*( *(chara+1) +2) );
/* int 指针数组转二级指针 */
int b1[] = {1,2,3};
int b2[] = {4,5,6};
int b3[] = {7,8,9};
int *intb [3] = {b1, b2, b3};
int **pob = intb;
funInt2OrderPoint(pob);
printf("\n");
printf("如果打开评论，可以执行，但是会报警告，结果是对应元素的地址\n");
// printf("%d\n", *intb+2 ); //&3
// printf("%d\n", *intb+1 ); //&2
// printf("%d\n", *intb ); //&1
// printf("\n");
// printf("%d\n", intb[0] ); //&123
// printf("%d\n", intb[1] ); //&456
// printf("%d\n", intb[2] ); //&789
printf("\n");
printf("%d\n", intb[1][2] );
printf("%d\n",*( intb[1] +2) );
printf("%d\n",*( *(intb+1) +2) );
return 0;
}
Summary
1 3 2 4
Form char *argv[] int *a3[] int *a2 int a4[][2] 传递时时参 argv a3 &a2 a4 接收时形参 char **p1 int **p3 int **p2 int(*p4)[2] argv[1][2] a3[0][1] a2[2] a4[1][0] p1[1][2] p3[0][1] p2[0][2] p4[1][0]
\*(\*(p+1)+2) \*(\*(p+0)+1) \*(\*(p+0)+2) \*(\*(p+1)+0)
#include <stdio.h>
int main(int argc, char *argv[])
{
char **p1 = argv;
printf("%c \n", argv[1][2] );
printf("%c \n", p1[1][2] );
printf("%c \n", *(*(p1+1)+2) );
int b2[] = {1, 2, 3, 4, 5};
int *a2 = b2;
int **p2 = &a2;
printf("%d \n", a2[1] );
printf("%d \n", *(a2+1) );
printf("%d \n", p2[0][1] );
printf("%d \n", *(*(p2+0)+1) );
int m[] = {1,2};
int n[] = {3,4};
int *a3[] = {m, n};
int **p3 = a3;
printf("%d \n", a3[0][1] );
printf("%d \n", p3[0][1] );
printf("%d \n", *(*(p3+0)+1) );
return 0;
int a4[][2] ={{1,2},{3,4}};
int (*p4)[2] = a4;
printf("%d \n", a4[1][0] );
printf("%d \n", p4[1][0] );
printf("%d \n", *(*(p4+1)+0) );
}
Polymorphism
C++ 先生成基类实列后再生成子类实例并指向
需要借助虚函数来实现对相应多态函数的调用，在函数前加上virtual
#include <iostream>
using namespace std;
class Animal
{
public:
Animal();
virtual run();
};
Animal::Animal(){};
void Animal::run(){cout <<"Animal is running"<<endl;}
class Dog : public Animal
{
public:
Dog();
virtual run();
};
Dog::Dog(){};
void Dog::run(){cout <<"Dog is running"<<endl;}
Animal *p = new Animal;
p->run();
p = new Dog;
p->run();
//此时的p就是多态变量，但是只有这个变量先由基类生成，再指向子类
//反之时不能的
Python 可基可子
class Animal(object):
def __init__(self):
self.name ='Animal name'
def run(self):
print('Animal is running')
class Dog(Animal):
def __init__(self):
self.name ='Dog name'
def run(self):
print('Dog is running')
# Polymorphism, all Class or instance will be checked the best passing
# mothode or character
def run_twice(a):
a.run()
def name(b):
print(b.name)
ani = Animal()
ani.name
ani.run()
ani = Dog()
ani.name
ani.run()
## ani 可以先有基类实现再多态映射到子类，也可以反向实现
Python 3.7.4 (default, Aug 13 2019, 20:35:49)
[GCC 7.3.0] :: Anaconda, Inc. on linux
Type "help", "copyright", "credits" or "license" for more information.
Animal is running
Dog is running
Cat is running
Dog is running
Animal name
Dog name
Dog name
python.el: native completion setup loaded
Java 子类实列指向父类引用
在向上转型后，就可以调用在所有子类中的同名函数
class Figure {
double dim1;
double dim2;
Figure(double d1, double d2) {
// 有参的构造方法
this.dim1 = d1;
this.dim2 = d2;
}
double area() {
// 用于计算对象的面积
System.out.println("父类中计算对象面积的方法，没有实际意义，需要在子类中重写。");
return 0;
}
}
class Rectangle extends Figure {
Rectangle(double d1, double d2) {
super(d1, d2);
}
double area() {
System.out.println("长方形的面积：");
return super.dim1 * super.dim2;
}
}
class Triangle extends Figure {
Triangle(double d1, double d2) {
super(d1, d2);
}
double area() {
System.out.println("三角形的面积：");
return super.dim1 * super.dim2 / 2;
}
}
public class Test {
public static void main(String[] args) {
Figure figure = new Rectangle(9, 9);
System.out.println(figure.area());
System.out.println("===============================");
figure = new Triangle(6, 8);
System.out.println(figure.area());
System.out.println("===============================");
figure = new Figure(10, 10);
System.out.println(figure.area());
}
}
Scanf
input matching for multi input Scanf()
可以直接从控制台接受8， 10， 16 进制的数 只有当控制字符串以格式控制符(%d、%c、%f) 开头时，键入的input才会忽略换行符, 否则输入的空白符就不能忽略了，它会参与匹配过程
清空每次输入的所有缓存
scanf("%*[^\n{=latex}]"); scanf("%*c");
Scanf(%{*}{width}type)
其中，{ } 表示可有可无。各个部分的具体含义是：
type表示读取什么类型的数据，例如 %d、%s、%[a-z]、%[^\n{=latex}]
等；type 必须有。 width表示最大读取宽度，可有可无。
*表示丢弃读取到的数据，可有可无。
hits
malloc for 2 diamesion
char ** commands; commands = (char **)malloc(NUMOFCOM * sizeof(char *)); for (int n = 0; n < NUMOFCOM; n++) commands[n] = (char *)malloc(sizeof(char) * COMLONG);
reference
1. 如果在函数体中修改了形参的数据，那么实参的数据也会被修改，从而拥有“在
函数内部影响函数外部数据”的效果
2. 不能返回局部数据（例如局部变量、局部对象、局部数组等）的引用，因为
当函数调用完成后局部数据就会被销毁，有可能在下次使用时数据就不存在
了
3. 给引用添加 const 限定后，不但可以将引用绑定到临时数据，还可以将引用
绑定到类型相近的数据，这使得引用更加灵活和通用，它们背后的机制都是
临时变量
class
1. 基类中的 protected 成员可以在派生类中使用，而基类中的 private 成员不能
在派生类中使用
2. 只有一个作用域内的同名函数才具有重载关系，不同作用域内的同名函数是
会造成遮蔽
3. 如果基类的成员变量被派生类的成员变量遮蔽， 基类成员仍会在实例化时被
创建，也可通过域解析符来访问。
4. 构造函数会被逐级的显示或者默认的在派生类中被调用，并且可以被重载
5. C++ 可以多继承不同的类（多继承和重继承），注意其构造函数的调用和成
员名称冲突，但是可以用域解析符来指明调用
6. 即使是类的private成员，仍能通过创建的对象的地址偏移或者直接利用指针
进行访问。（有点花里胡哨的）
7. C++中虚函数的唯一用处就是构成多态
8. 引用不像指针灵活，指针可以随时改变指向，而引用只能指代固定的对象
const
const 的作用在C++中和宏很像#defind
函数中const，修饰变量后可以将传入的参数，强制转换为设定的
初始化 const 成员变量的唯一方法就是使用参数初始化
类中的const， 变量，函数， 类， 只能互相承
重要：在头文件中用const修饰全部变量后，就可以多次被引入了而不会出现重
复定义的错
static
statis 变量， 函数，全部对象可以共用，访问，无this指针。不能调用普通变量和函数类
static 多被用来计数，可以在外部被改
GDB
can be done in terminal or in Emacs, recommend later one
b add break point run start the program run argv if argv needed n next l list source code 10 lines p print variables s go into functions ignore pass the break point q exit set set variable
in emacs, if scanf, the input can not be given in gdb buffer. if it comes to scanf, go to the I/O buffer, and input the value , go back to gdb buffer, just next
Compile
There are four different kinds of Methods to compile a source file.
Makefile
There is a folder called Makefile, into this folder, and call "make" in terminal, don't forget "make clean" to clean it.
Terminal Compile
In pthread~andpid~ folder with terminal with "g++/gcc file.c -o file", and then "./file" to call it
Emacs Compile
Also in pthread~andpid~ folder, open the resource code with Emacs, and then M-x compile(C-z k) call it in minibuffer with : "gcc -pthread create~pthread~.c -o create~pthread~ && ./create~pthread~"
Literatur programmierung
pass the agruments to program
在src block 中提前定义:var a = 3
javac test.java && echo 1 |java test 可以将1向StdInt传入
gcc test.c && echo 1 | ./a.out 可以将1向scanf("%d", &a)传入的a
Beispile
#+header: :var input=23 :var b1 = 0 :var b2=1 :var b3=3
int b[] = {b1, b2, b3};
printf("niiiiilhakkko\n");
printf("%d\n", input);
printf("%d\n", b2);
for(int i = 0; i<3;i++){
printf("%d", b[i]);
}
niiiiilhakkko
23
1
013