Question

Assume x is a variable of type pointer (i.e., an address) that points to onebyte unsigned number. Further, assume b is a variable of size 4 bytes and is stored

at memory address 0xA80.
Suppose these variables are given values as follows.
b = 0x2F552;
x = 0xA81;
Hint: This problem is about how to write data in memory with different endianness.
1.1 Suppose the system is big-endian. Draw a diagram (as an array, with addresses
as indices) of memory space where b is stored and their contents, and depict x as
a pointer (arrow) to the appropriate byte of b. What is the (hexadecimal) value
that x points to?
1.2 Now suppose the system is little-endian. Draw a new picture, contrasting with
part (a). What is the (hexadecimal) value that x points to?

Answer 1

In a big-endian system, x points to 0x2F. In a little-endian system, x points to 0x2F52. In a big-endian system, the variable x would point to the value 0x2F at address 0xA81. For a little-endian system, x would point to the value 0x55 at the same address. Endianess affects how data is stored across memory addresses.

Step-by-step explanation:

In a big-endian system, the most significant byte is stored at the lowest memory address. So, in this case, the memory space for variable b would look like this:

Address Contents
0xA80 0x00
0xA81 0x00
0xA82 0x00
0xA83 0x2F

The value that x points to in this case is 0x2F.

In a little-endian system, the least significant byte is stored at the lowest memory address. So, the memory space for variable b would look like this:

Address Contents
0xA80 0x52
0xA81 0x2F
0xA82 0x05
0xA83 0x00

The value that x points to in this case is 0x2F52.

In a big-endian system, the variable x would point to the value 0x2F at address 0xA81. For a little-endian system, x would point to the value 0x55 at the same address. Endianess affects how data is stored across memory addresses.

The question pertains to how data is stored in memory with respect to endianess. Endianess determines the order in which bytes are arranged within larger data types. In a big-endian system, the most significant byte (MSB) is stored at the smallest address, and in a little-endian system, the least significant byte (LSB) is stored at the smallest address. Given that the variable b has the hexadecimal value 0x2F552 and is 4 bytes in size, its representation in memory will differ depending on the system's endianess.

In a big-endian system, the value of b will be stored across four bytes starting from address 0xA80. Thus, the memory representation will be as follows:

0xA80: 00 (since 0x2F552 fits within 3 bytes, the MSB is 0)

0xA81: 2F

0xA82: 55

0xA83: 52

Since x points to the address 0xA81, the value that x points to will be 0x2F.

In contrast, a little-endian system reverses the byte order. The memory representation would then be:

0xA80: 52

0xA81: 55

0xA82: 2F

0xA83: 00

Here, since x points to address 0xA81, the value that x points to will be 0x55.