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CSE238-Project 1 Data Type Converter Solved
The purpose of this assignment is to become more familiar with bit-level representations of integers and floating point numbers.
In this project, you will implement an application, in C or Java programming language; that takes
• a file containing signed integers, unsigned integers and floating point numbers
• the byte ordering type (Little Endian or Big Endian),
• the size of the floating point data type (the size of both signed and unsigned integers are 2 bytes), as input and converts the contents of the file to hexadecimal according to the predefined format and gives the converted data as output.
The size of the data type can be 1, 2, 3, or 4 bytes.
• The signed integers are written as normal integers in the file. (e.g., 3, -5, etc.)
• The unsigned integers are written as integers followed by the letter u. (e.g., 3u, 5u, etc.)
• The floating point numbers are written with a decimal point. (e.g., 3.0, 5.987, etc.)
• If the read data type is signed integer, your program will convert the numbers in the input file using 2’s complement representation.
• If the read data type is unsigned integer, numbers will be converted using unsigned integer representation.
• If the read type is floating point number, you will use IEEE-like format. The number of exponent bits according to given data size will be like the following:
o if 1 byte (i.e., 8 bits), 4 bits will be used for exponent part o if 2 bytes (i.e., 16 bits), 6 bits will be used for exponent part o if 3 bytes (i.e., 24 bits), 8 bits will be used for exponent part o if 4 bytes (i.e., 32 bits), 10 bits will be used for exponent part
o While calculating the mantissa to get the floating point value, you will only use the first 13 bits of the fraction part (for 3-byte and 4-byte data sizes). You will use “round to nearest even” method for rounding fraction bits to 13 bits.
Details about the program are listed below:
• At the beginning of the execution, your program will prompt for the input file.
• In the input file, each line will include a single number. Example: input.txt
29.109375
4u
-9
6u
0
-63.0
• After a valid input file is taken as input, the user will be prompted for the byte ordering type and size:
Example:
Byte ordering: Little Endian
Floating point size: 2 bytes
• You program will calculate the hexadecimal value of the file content with the given information:
Example:
The byte ordering is Little Endian, and the floating point is 2 byte.
o First, your program will read the first number of the file. For our “input.txt” the first number is:
29.109375
o This is a floating point number.
29.109375 = 11101.000111 = 1.1101000111 * 24
o Mantissa is 1.1101000111. However, for 2 bytes, we have 9 bits of fraction part, so we need to round to nearest even: 1.1101000111 ≈ 1.110100100. The fraction part is 110100100
o E = 4. Exponent is E = exp – Bias, where Bias = 26-1 – 1 = 31. Therefore the equation is 4 = exp - 31. exp = 35 which is 100011 in binary.
o The number is positive, so sign bit is 0. o The number at total is 0100011110100100 which is 0x47A4 in hexadecimal.
o The byte ordering is Little Endian, so the floating point number is represented as:
A4 47
In the output file, the printed value will be: A4 47 Example 2:
o If the program reads the following number, it is:
4u
o This is an unsigned integer. Unsigned integers are 2 bytes:
4 = 0000000000000100
o 0000000000000100 is 0x0004 in hexadecimal.
o The byte ordering is Little Endian, so the unsigned integer is represented as:
04 00
In the output file, the printed value will be: 04 00
• The output file will be:
output.txt
A4 47
04 00
F7 FF
06 00
00 00 F0 C9
• You cannot use library functions for the decimal to binary conversions.
