CPP Programming Language

Generic Operators §

For C++,

 
#include<iostream>
using namespace std;
 
// Represents the main function that is ran when file is compiled
int main(){
	cout << "BRUH" << endl;
	// <<   -> write
	// >>   -> read
	// endl -> new line
	// `\n` -> new line
	// ;    -> Ends the line of code
	return 0;
};

Data Types and Variables §

• For declaring a variable
• [DATATYPE] [NAME] = [VALUE]

Data Type	Meaning	Size (in Bytes)
int	Integer	2 or 4
float	Floating-point	4
double	Double Floating-point	8
char	Character	1
wchar_t	Wide Character	2
bool	Boolean	1
void	Empty	0

• Char can only store 1 character
• The values are stored in terms of bits
  • Numbers in binary
  • Characters according to ASCII -> Binary

Typecasting §

One type can be converted to another using this method

#include<iostream> 
using namespace std;
int main(){
	int a = 'a'; // Converts it to the respective Denary value according to ASCII table
	char b = 97;
	cout << "The ASCII value of a is :" << a << endl;
	cout << "The char value of 97 is :" << b << endl;
	
	// Successfully ending a program
	return 0;
}

Due to the difference in sizes for the datatypes, It is possible to overflow a datatype, In which case it uses the final x amount of byte where x depicts the amount of bytes that data type can store For example char stores 1 byte, so

• max val = $2^8-1$
• min val = 0 For an int stores 4 bytes, so
• max val = $2^{32}-1$
• min val = 0

int main(){
	char a = 4952; 
	cout << a << endl;
	return 0;
}

An error is observed: In file included from input_line_3:1: /tmp/temp_1692542671892.cpp:7:11: warning: implicit conversion from 'int' to 'char' changes value from 4952 to 88 [-Wconstant-conversion] char a = 4952;

Our number 4952 in terms of binary is: 10011 01011000 The last 8 bits (size of char) make up the value of 88, Which in ASCII means X

How Negative Numbers are stored §

• 1st bit denotes the sign
  • +ve = 0
  • -ve = 1
• -5
  • Ignore -ve sign and convert to binary 101
  • Take 2’s compliment and store %%🖋 Edit in Excalidraw, and the dark exported image%%

The value is stored in the compliment form, And when printed. Stored numbers 2’s compliment is taken again (Converting it into a normal binary form), And the first bit of the original data gives the sign

This helps in saving an additional slot for a useless value as $-0=0$ If the first bit always represented the numbers sign, It would be wasted in this case Hence 2’s compliment is used The range for it starts from $-2^{31}<=x<=2^{31}-1$ instead of starting from $-2^{31}-1$ This can be “first bit for sign” can be ignored by using the unsigned operator

int main(){
	unsigned int a = 23; // A unsigned +ve number
	unsigned int b = -23; // A unsigned -ve number
	cout << a << endl;
	cout << b << endl;
	return 0;
}

• Note that the negative number is huge because the first value is taken to be RELEVANT/Unsigned

Operator §

Operator	Operation
+	Addition
-	Subtraction
*	Multiplication
/	Division
%	Modulo Operation (Remainder after division)

Operator	Example	Equivalent to
=	a = b;	a = b;
+=	a += b;	a = a + b;
-=	a -= b;	a = a - b;
*=	a *= b;	a = a * b;
/=	a /= b;	a = a / b;
%=	a %= b;	a = a % b;

Operator	Meaning	Example
==	Is Equal To	3 == 5 gives us false
!=	Not Equal To	3 != 5 gives us true
>	Greater Than	3 > 5 gives us false
<	Less Than	3 < 5 gives us true
>=	Greater Than or Equal To	3 >= 5 give us false
<=	Less Than or Equal To	3 <= 5 gives us true

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