# Bitwise Operation

## Lecture PPT

{% file src="/files/FEAdSADQhj0qWEcqWigM" %}

## Online Lesson

**코딩도장 핵심요약**: [비트연산자 사용하기 핵심요약](https://dojang.io/mod/page/view.php?id=490)

### Bitwise Operation in C

자료형과 메모리 주소를 바이트 단위로 구분하여 사용하였습니다. 비트 연산자는 **바이트** 단위보다 더 작은 **비트** 단위로 연산하는 연산자입니다.

![](/files/-MhCIegeANTqXDwl_QXK)

<figure><img src="/files/htb0nVRMgeAioa3sFj4w" alt=""><figcaption></figcaption></figure>

#### Example

```cpp
	PA=PA | (1<<5);  	// set PA5 (as High) and mask others
	PA=PA & ~(1<<5);  	// reset PA5 (as LOW) and mask others
	bit = PA & (1<<5);  // check the bit5. bit=1 if  PA5 is 1
```

###

***

### Set flag: (**플래그 |= 마스크)**

> a |= (1 << k)

![](/files/-MhCLFrTyxgpUPM3XiTV)

### Clear flag (**플래그 &= \~마스크)**

Example:

```cpp
unsigned char flag = 7;    // 7: 0000 0111
flag &= ~2;    

0000 0010
_________ ~
1111 1101
```

![](/files/-MhCLQ_q9Odc_RBhmwsM)

###

### Toggle flag **(플래그 ^= 마스크)**

> a ^= 1<\<k

![](/files/-MhCMOJeY_Ytw785govL)

### Read a bit

> (Method 1) bit = a & (1<\<k) // Shift ‘bit 1’ left by k starting from LSB
>
> (Method 2) bit = (a >>k) & (1) // Shift target ‘bit right by k

Example:

<figure><img src="/files/nLZKFTcBg4puFVvz2DCa" alt=""><figcaption></figcaption></figure>

### Read multiple bits

<figure><img src="/files/QfKkZigqZhrlT5OKVDDE" alt=""><figcaption></figcaption></figure>

### **Tip: Use Macro**

```cpp
#define BIT_SET(REG, BIT)		((REG) |= 1<< (BIT))
#define BIT_CLEAR(REG, BIT)		((REG) &= ~1<<(BIT))
#define BIT_READ(REG, BIT)		((REG)>>BIT & (1))
#define BITS_CLEAR(REG, BIT,NUM)	((REG) &= ~((0x1<< NUM)-1)<<(BIT))
#define BITS_SET(REG, BIT,NUM)		((REG) |= NUM<< (BIT))
```

##

***

## Exercise

### Exercise 1

What will be the output ?

* [C\_bitwise\_exercise\_01.c](https://github.com/ykkimhgu/Tutorial-C-Program/blob/main/bitwise/EC_bitwise_exercise_01.c)

{% tabs %}
{% tab title="Exercise" %}

```cpp
#include <stdio.h>
 
int main()
{
    unsigned char num1 = 1;    // 0000 0001
    unsigned char num2 = 3;    // 0000 0011
    unsigned char num3 = 162;    // 162: 1010 0010
    unsigned char num4;
    num4 = ~num3;

 
    printf("%d\n", num1 & num2);    
    printf("%d\n", num1 | num2);    
    printf("%d\n", num1 ^ num2);    
    printf("%u\n", num4);    // 93: 0101 1101: num1의 비트 값을 뒤집음
 
    return 0;
}
```

{% endtab %}

{% tab title="Solution" %}

```cpp
#include <stdio.h>
 
int main()
{
    unsigned char num1 = 1;    // 0000 0001
    unsigned char num2 = 3;    // 0000 0011
    unsigned char num3 = 162;  // 162: 1010 0010
    unsigned char num4;
    num4 = ~num3;

    printf("%d\n", num1 & num2);    // 0000 0001: 01과 11을 비트 AND하면 01이 됨
    printf("%d\n", num1 | num2);    // 0000 0011: 01과 11을 비트 OR하면 11이 됨
    printf("%d\n", num1 ^ num2);    // 0000 0010: 01과 11을 비트 XOR하면 10이 됨
    printf("%u\n", num4);    // 93: 0101 1101: num1의 비트 값을 뒤집음
 
    return 0;
}
```

{% endtab %}
{% endtabs %}

###

### Exercise 2

What will be the output?

* [C\_bitwise\_exercise\_02.c](https://github.com/ykkimhgu/Tutorial-C-Program/blob/main/bitwise/EC_bitwise_exercise_02.c)

{% tabs %}
{% tab title="Exercise" %}

```cpp
#include <stdio.h> 

int main()
{
    unsigned char num1 = 4;    // 4: 0000 0100
    unsigned char num2 = 4;    // 4: 0000 0100
    unsigned char num3 = 4;    // 4: 0000 0100
    unsigned char num4 = 4;    // 4: 0000 0100
    unsigned char num5 = 4;    // 4: 0000 0100
 
    num1 &= 5;     // 5(0000 0101) AND 연산 후 할당
    num2 |= 2;     // 2(0000 0010) OR 연산 후 할당
    num3 ^= 3;     // 3(0000 0011) XOR 연산 후 할당
    num4 <<= 2;    // 비트를 왼쪽으로 2번 이동한 후 할당
    num5 >>= 2;    // 비트를 오른쪽으로 2번 이동한 후 할당
 
    printf("%u\n", num1);    
    printf("%u\n", num2);    
    printf("%u\n", num3);    
    printf("%u\n", num4);    
    printf("%u\n", num5);    
 
    return 0;
}
```

{% endtab %}

{% tab title="Solution" %}

```cpp
#include <stdio.h> 

int main()
{
    unsigned char num1 = 4;    // 4: 0000 0100
    unsigned char num2 = 4;    // 4: 0000 0100
    unsigned char num3 = 4;    // 4: 0000 0100
    unsigned char num4 = 4;    // 4: 0000 0100
    unsigned char num5 = 4;    // 4: 0000 0100
 
    num1 &= 5;     // 5(0000 0101) AND 연산 후 할당
    num2 |= 2;     // 2(0000 0010) OR 연산 후 할당
    num3 ^= 3;     // 3(0000 0011) XOR 연산 후 할당
    num4 <<= 2;    // 비트를 왼쪽으로 2번 이동한 후 할당
    num5 >>= 2;    // 비트를 오른쪽으로 2번 이동한 후 할당
 
    printf("%u\n", num1);    //  4: 0000 0100: 100과 101을 비트 AND하면 100이 됨
    printf("%u\n", num2);    //  6: 0000 0110: 100과 010을 비트 OR하면 110이 됨
    printf("%u\n", num3);    //  7: 0000 0111: 100과 011을 비트 XOR하면 111이 됨
    printf("%u\n", num4);    // 16: 0001 0000: 100을 왼쪽으로 2번 이동하면 10000이 됨
    printf("%u\n", num5);    //  1: 0000 0001: 100을 오른쪽으로 2번 이동하면 1이 됨
 
    return 0;
}
```

{% endtab %}
{% endtabs %}

### Exercise 3

Download and Read instruction in the given source file.

Fill in the blanks.

* [C\_bitwise\_exercise\_03.c](https://github.com/ykkimhgu/Tutorial-C-Program/blob/main/bitwise/EC_bitwise_exercise_03.c)

{% tabs %}
{% tab title="Exercise" %}

```cpp
#include <stdio.h>
#include <stdint.h>

void dec2bin(unsigned int n);

void main() {
    
	uint8_t vals=0;
	
	// Assume 8 LEDs are connected to Digital Out pins of Port A (PA)
	// LED0 at PA[0] to LED7 at PA[7]
	uint8_t PA;
	
	// Initial Values of PA
	PA = 0b00001111;			
	printf("\n Initial PA: ");
	dec2bin(PA);
	
	
	// Exercise_1-1: Turning ON LED4 at PA[4] 
	PA |=  ________; 	    // turn ON LED4 
	printf("\n ex1-1: ");
	dec2bin(PA);
	
	//Exercise_1-2:  Read the bit at PA[4], 4th from LSB
	vals =  ________;	// read bit of PA[4]
	printf("\n ex1-2: %d\n", vals);
	
	
	// Exercise_2-1: Turn off LED4 at PA[4]
	PA &= ________;    	// turn off LED4  
	printf("\n ex2-1: ");
	dec2bin(PA);
	
	//Exercise_2-2:  Read the bit at PA[4], 4th from LSB
	vals =  ________;	// read bit of PA[4]
	printf("\n ex2-2: %d\n", vals);
	
	
	
	// Exercise_3-1: Turning ON LED5 and LED4,  at PA[5:4] 
	PA |=  ________;	// turn ON LED5 and LED4
	printf("\n ex3-1: ");
	dec2bin(PA);
	
	//Exercise_3-2:  Read bits LED5 and LED4,  at PA[5:4] 
	vals =  ________;	// read bits at PA[5:4] 
	printf("\n ex3-2: %d\n", vals);
	
	//Exercise_4-1:  SET  PA[0] and PA[7] 
	PA |= 0b10000001; 
	printf("\n ex4-1: ");
	dec2bin(PA);
	
	//Exercise_4-2:  CLEAR  PA[0] and PA[7] 
	PA &=  ________;
	printf("\n ex4-2: ");
	dec2bin(PA);
	
	
	//Exercise_5-1:  Toggle LED7 PA[7]
	printf("\n");    
	PA ^=  ________;
	printf("\n ex5-1: ");
	dec2bin(PA);
	
	printf("\n");    
	
}

void dec2bin(unsigned int n) {
	unsigned int a = 0x80;
	for (int i = 0; i < 8; i++) {
		if ((a & n) == a)
			printf("1");
		else
			printf("0");
		a = a >> 1;
	}
}
```

{% endtab %}

{% tab title="Outut" %}

```
 PA   : 00001111
 ex1-1: 00011111
 ex1-2: 1

 ex2-1: 00001111
 ex2-2: 0

 ex3-1: 00111111
 ex3-2: 3

 ex4-1: 10111111
 ex4-2: 00111110

 ex5-1: 10111110

```

{% endtab %}
{% endtabs %}

###


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