//5-1-1 math symbols
#include <iostream>
using namespace std;
int main() {
int a = 1;
int b = 3;
cout << a+b << endl;
cout << a – b << endl;
cout << a * b << endl;
cout << a / b << endl; //if int divide int, there will be int too
cout << a*1.0 / b << endl;//if int multiple float before divide, there will be float
a = 10000000;
b = 100000;
cout << a * b << endl; //output will out the range of int
cout << (long long)a * b << endl; //we can translate it forcely
a = -1;
b = 2;
cout << a / b << endl;//just cut of the dot number
a = +1;
b = -a;
int c = -(-a);
cout << a <<‘ ‘ << b <<‘ ‘ << c << endl;//postive & negative
char A = ‘A’;
A = A + 1;
A = A + 24;
cout << A << endl;
return 0;
}
//5-1-2 mod % symbol
#include <iostream>
using namespace std;
int main(){
int a = 100;
int b = 9;
cout << a % b << endl; //1
a = 100;
b = -9;
cout << a % b << endl; //1
a = -100;
b = 9;
cout << a % b << endl;//the mod symbol %;-100 % 9 is 9*-11 (near 0 direction) not -12 in C/C++ launguage
//-1
a = -100;
b = -9;
cout << a % b << endl; //-100 -(-9)*11=-1
//-1
//1、the symbol of mod is the same as the be devided number
//first we should take the negative symbol from the be devided number
//second we use it as 100 = -9 * （-11） + 1
//third the left +1 is the lease number and add the negative symbol – to the 1
//fourth the final answer is -1
return 0;
}
//5 – 1-3 ++ & –symbol
#include <iostream>
using namespace std;
int main() {
int a = 6;
a++; //a=a+1
cout << a << endl;
++a; //a=a+1
cout << a << endl;
//the same in the two ways
int j = 8;
int x = a++; //first give the value to x, then add 1 to a
int y = ++j; //first add 1 to the value , then give the value to y (this type is more effective)
cout << x << endl;
cout << y << endl;
int z = (a++) + (++a);//9+11=20 we must not to write this type code ,because it will make a big problem
cout << z << endl;
a–; //a=a-1
cout << a << endl;
–a;//a=a-1
cout << a << endl;
}
//5 – 2 valued symbols
#include <iostream>
using namespace std;
int main() {
int x = 9;
int y = 6;
x = y; // let y to x
cout << x << endl;
x += y; //x=x+y the length is 10 bytes to 7 bytes
cout << x << endl;
x -= y; //x=x-y
cout << x << endl;
x *= y; //x=x*y
cout << x << endl;
x /= y; //x=x/y
cout << x << endl;
x %= y; //x=x%y
cout << x << endl;
return 0;
}