Decimal to binary advanced version (can convert negative numbers and boundary values)

Decimal to binary

Ten Base number Negative to binary

Let's say I have a int Number of types , The value is 5, that , We know that it's represented in a computer as ： ( because java in int yes 4 Bytes , So the high position needs to be filled 0, Occupy enough 32 position )

00000000 00000000 00000000 00000101

Now I want to know ,-5 How to express in a computer ？

In the computer , Negative numbers begin with Original code In the form of complement .

What do you mean Complement code Well ？ This has to come from the original , The irony is that .

Original code ： A positive number , Converted to... According to the absolute value Binary system Count ; A negative number , Binary number converted by absolute value size , And then the highest bit makes up for 1, It's called the original code .

such as ：

5 The original code of ： 00000000 00000000 00000000 00000101

-5 The original code of ： 10000000 00000000 00000000 00000101

Inverse code ： The inverse of a positive number is the same as the original , The inverse of a negative number is the inverse of the original code of the number except for the sign bit .

Reverse operation means ： originally 1, have to 0; originally 0, have to 1.(1 change 0; 0 change 1)

such as ：

Positive numbers : 00000000 00000000 00000000 00000101 It's the inverse of 00000000 00000000 00000000 00000101 ;

negative : 10000000 00000000 00000000 00000101 Reverse each bit ( Except the sign bit ), have to 11111111 11111111 11111111 11111010.

call ：10000000 00000000 00000000 00000101 and 11111111 11111111 11111111 11111010 It's the opposite of each other .

Complement code ： The complement of a positive number is the same as the original code , The complement of a negative number is to invert the original code of the number except the sign bit , Then add... To the last 1.

such as ：

-5 The original code of ：10000000 00000000 00000000 00000101

-5 The inverse of ：11111111 11111111 11111111 11111010

-5 Complement ：11111111 11111111 11111111 11111010 + 1 = 11111111 11111111 11111111 11111011

therefore ,-5 It's expressed in a computer as ：11111111 11111111 11111111 11111011. Convert to hex ：0xFFFFFFFB.

Take another example , Let's look at integers -1 How to express in a computer ：

Suppose this is also a int type , that ：

1、 First -1 The original code of ：10000000 00000000 00000000 00000001

2、 We have to counter ： 11111111 11111111 11111111 11111110( The division sign bit is reversed bit by bit )

3、 Get complement ： 11111111 11111111 11111111 11111111

#include <stdio.h>

int main()
{
	while(1)
	{
		int i,count=0,j=0;
		long long n=0;
		int a[64]={0};
		printf("\n");
		printf(" Please enter a decimal integer :\n");
		scanf("%lld",&n);
		if(n>=0)
		{
			while(n!=0)
			{
				a[count]=n%2;
				n=n/2;
				count++;
			}
			printf(" The value is converted to binary number ：");
			for(i=63;i>=0;i--)
			{
				if((i+1) % 4 == 0 && i != 63)	
					printf(" ");
				printf("%d",a[i]);
			}
		}
		else 
		{
			n++;
			n=-n;// Take the absolute value first  
			while(n!=0)
			{
				a[count]=n%2;
				n=n/2;
				count++;
			}
			for(i=0;i<64;i++)// Take the inverse ,0 change 1,1 change 0
			{    
				if(a[i]==0)
					a[i]=1;
				else
					a[i]=0;
			}
			printf(" The value is converted to binary number ：");
			for(i=63;i>=0;i--)
			{
				if((i+1) % 4 == 0 && i != 63)
					printf(" ");
				printf("%d",a[i]);
			}		
		}	
	}
	return 0;
}

  Running results ：

  Employee ranking ：

#include <stdio.h>
int count(int arr_08[7],int x);

int main()
{
	int i,j,sum=0,temp=0,arr_02[2][8],arr[8][7]={
   {2,4,3,4,5,8,8},
								{7,3,4,3,3,4,4},
								{3,3,4,3,3,2,2},
								{9,3,4,7,3,4,1},
								{3,5,4,3,6,3,8},
								{3,4,4,6,3,4,4},
								{3,7,4,8,3,8,4},
								{6,3,5,9,2,7,9}};
	printf(" Statistics of working hours of each employee ：\n");
	for(i=0;i<8;i++)// Print work duration  
	{
		for(j=0;j<7;j++)
		{
			printf("%d  ",arr[i][j]);
		}
		printf("\n");
	}
	printf(" The total working hours of each employee are ：\n");
	for(i=0;i<8;i++)
	{
		sum=0;
		for(j=0;j<7;j++)
			sum += arr[i][j];
		printf("%d\t",sum);
		arr_02[0][i]=sum;
		arr_02[1][i]=i;
	}	
	printf("\n");
	for(i=0;i<2;i++)
	{
		for(j=0;j<8;j++)
			printf("%d\t",arr_02[i][j]);
		printf("\n");
	}
	for(i=0;i<8;i++)
	{
		for(j=0;j<8-i-1;j++)
		{
			if(arr_02[0][j]<arr_02[0][j+1])
			{
				temp=arr_02[0][j];
				arr_02[0][j]=arr_02[0][j+1];
				arr_02[0][j+1]=temp;
				
				temp=arr_02[1][j];
				arr_02[1][j]=arr_02[1][j+1];
				arr_02[1][j+1]=temp;
			}
		}
	}
	printf(" The total working hours of each employee are in descending order ：\n");
		for(j=0;j<8;j++)
			printf("%d\t",arr_02[0][j]);	
	printf("\n");
	printf(" The total duration of each employee's employee number is in descending order ：\n");
		for(j=0;j<8;j++)
			printf("%d\t",arr_02[1][j]);	
	return 0;
}

  Running results ：