In a C program, when we use the following statements:
int a = 12;
The operating system allocates 4 bytes of memory for a and stores 12 in binary format in those 4 bytes.
In binary 12 = 1100, which when expanded to store in 4 bytes will be as follows:
00000000 00000000 00000000 00001100.
Lets assume the memory address to store these 4 bytes are 1001,1002,1003,1004
The rightmost bit is called the least significant bit and the leftmost bit is the most significant bit. Each byte is stored in a different memory address. If we store the bytes with the least significant byte in the lowest memory address ie store 00001100 in 1001, this method of storing in memory is called little endian. On the other hand, if we store the most significant byte in the lowest memory address ie store 00000000 in 1001, it is called big endian.
Integers can be positive or negative. One way to store negative numbers in binary is to set the most significant bit to 1 to denote that it is a negative number. So -12 is stored as follows:
10000000 00000000 00000000 00001100.
So, we have only 31 bits for the numbers. Thus the largest integer that can be stored is 2^31-1= 65535 and the lowest integer that can be stored is -2^31-1= -65535
However, there are two major problems when we store integers like this.
1. Lets suppose we want to store values from -3 to +3. There are 7 numbers in this range.
0 - 000
1 - 001
2 - 010
3 - 011
-0 - 100
-1 - 101
-2 - 110
-3 - 111
Here we can see that 0 has two representations which can create a big confusion.
2. Arithmetic operations do not give the correct results.
If we add +1 and -1, it should output 0. Here,
001 + 101 = 110 which is -2 and that is wrong.
So, to avoid these we use 2's complement to store negative numbers. In 2's complement, to get the binary representation for a negative number, its positive conterpart is first complemented and 1 is added to the 1's complement.
The 2's complement of 1(001) is 111(110+1).
The above range can now be represented as follows:
0 - 000
1 - 001
2 - 010
3 - 011
-1 - 111
-2 - 110
-3 - 101
-4 - 100
Thus, we can see that both of the above problems are solved using the 2's complement. Also, we can store one more number as there is only 1 representation for 0. The range of number which can be represented using this method is -2^31 to 2^31-1.
int a = 12;
The operating system allocates 4 bytes of memory for a and stores 12 in binary format in those 4 bytes.
In binary 12 = 1100, which when expanded to store in 4 bytes will be as follows:
00000000 00000000 00000000 00001100.
Lets assume the memory address to store these 4 bytes are 1001,1002,1003,1004
The rightmost bit is called the least significant bit and the leftmost bit is the most significant bit. Each byte is stored in a different memory address. If we store the bytes with the least significant byte in the lowest memory address ie store 00001100 in 1001, this method of storing in memory is called little endian. On the other hand, if we store the most significant byte in the lowest memory address ie store 00000000 in 1001, it is called big endian.
Integers can be positive or negative. One way to store negative numbers in binary is to set the most significant bit to 1 to denote that it is a negative number. So -12 is stored as follows:
10000000 00000000 00000000 00001100.
So, we have only 31 bits for the numbers. Thus the largest integer that can be stored is 2^31-1= 65535 and the lowest integer that can be stored is -2^31-1= -65535
However, there are two major problems when we store integers like this.
1. Lets suppose we want to store values from -3 to +3. There are 7 numbers in this range.
0 - 000
1 - 001
2 - 010
3 - 011
-0 - 100
-1 - 101
-2 - 110
-3 - 111
Here we can see that 0 has two representations which can create a big confusion.
2. Arithmetic operations do not give the correct results.
If we add +1 and -1, it should output 0. Here,
001 + 101 = 110 which is -2 and that is wrong.
So, to avoid these we use 2's complement to store negative numbers. In 2's complement, to get the binary representation for a negative number, its positive conterpart is first complemented and 1 is added to the 1's complement.
The 2's complement of 1(001) is 111(110+1).
The above range can now be represented as follows:
0 - 000
1 - 001
2 - 010
3 - 011
-1 - 111
-2 - 110
-3 - 101
-4 - 100
Thus, we can see that both of the above problems are solved using the 2's complement. Also, we can store one more number as there is only 1 representation for 0. The range of number which can be represented using this method is -2^31 to 2^31-1.
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