C辅导件I/O处理，数字的进制转换 Problems解析c编程、c调试

Requirement

You may discuss any of the assignments with your classmates and instructors (or anyone else) but all work for all assignments must be entirely your own.

Any sharing or copying of assignments will be considered cheating (this includes posting of partial or complete solutions on Piazza, GitHub public repositories or any other public forum). If you get significant help from anyone, you should acknowledge it in your submission (and your grade will be proportional to the part that you completed on your own).

You are responsible for every line in your program: you need to know what it does and why.

The Makefile provided with this project compiles and builds all of the required programs.
To compile and build all of the program run:

make all

To compile and build individual problems run:

make prob1
make prob2 
...

To run programs for each problem you can either execute

./prob1 in_file out_file 
./prob2
...

or, you can use the rules defined in the Makefile:

make run_prob1
make run_prob2
...

Finally, to run programs using valgrind you can use the rules
defined in the Makefile:

make test_prob1
make test_prob2
...

To remove all previously compiled and built files, run

make clean

Problem 1

The code in prob1.c is a slightly modified version of the code example from the manual page of getline() function.

Read the documentation of getline() function and answer the questions in problem1_answers.md file.

Deliverables

Answers to the questions in the file problem1_answers.md

Problem 2

Some nasty scrambler messed up the text of the nursery rhyme [Mary Had a Little Lamb] (https://en.wikipedia.org/wiki/Mary_Had_a_Little_Lamb) - the result is in the file called mary_corrupt.txt.

Your task for this problem is to “unscramble” the characters written to mary_corrupt.txt to recover the original nursery rhyme.

The nasty scrambler left some clues:

Clue 1: the corrupted file is still a text file - you should be able to work with individual ASCII characters even though they do not look like the English characters

Clue 2: for each original character, the 4 high order bits and the 4 low order bits were swapped.

Clue 3: the swap operation (when performed correctly) the second time should produce the original file back

Clue 4: the type char in C is signed, you may find it handy to use the unsigned char in parts of the solution.

Programming requirements

Implement the program in prob2.c file. All functions should be specified in that file.
The program should work with two command line arguments: the first one is the name of the input file, the second is the name of the output file.
The program should work with any “scrambled” input file that is created following the same pattern, not only with mary_corrupt.txt.
The program should be leak-free: any memory that is allocated should be freed before the program terminates.
The program has to be documented (file preamble with brief description and name of the author, inline comments).

Deliverables

Implementation of the program in file prob2.c.

Problem 3

The program in prob3.c extract information about bit-vector representation of six single precision floating point numbers. It uses several functions to accomplish that. Your job is to implement most of those functions.

The header file proj1.h provides the declarations and descriptions of those functions.

The definitions of those function should be written in proj1.c.

The program in prob3.c should serve as a guide for correctness of the implementation, but passing those tests does not guarantee the correctness of the implementation. We will test the function implementation on other values.

Programming requirements:

Implement the required functions in proj1.c file:

int is_normalized ( float f );
int is_denormalized ( float f );
int is_special ( float f );
char * get_M ( float f );
int get_s ( float f );
int get_E ( float f );

Implement any additional functions in proj1.c files. Add the declarations of those functions to proj1.h if need be.
The functions should work with any single precision floating point numbers.
The program should be leak-free: any memory that is allocated should be freed before the program terminates.
The code has to be documented (the author’s name in proj1.c, function descriptions, inline comments).
You should not use any of the functions defined in the math library.

Deliverables:

Implementation if functions in proj1.c file.

Problem 4

The program in prob4.c converts six bit-vectors to their single precision floating point equivalents. It uses several functions to accomplish that. Your job is to implement most of those functions.

The header file proj1.h provides the declarations and descriptions of those functions.

The definitions of those function should be written in proj1.c.

The program in prob4.c should serve as a guide for correctness of the implementation, but passing those tests does not guarantee the correctness of the implementation. We will test the function implementation on other values.

Programming requirements

Implement the required functions in proj1.c file:

float get_float_from_bits( char * bits );
type get_type_from_bits( char * bits );
int get_s_from_bits ( char * bits );
int get_E_from_bits ( char * bits );
float get_M_from_bits ( char * bits );

Implement any additional functions in proj1.c files. Add the declarations of those functions to proj1.h if need be.
The functions should work with any 32-bit long c-string sequences.
The program should be leak-free: any memory that is allocated should be freed before the program terminates.
The code has to be documented (the author’s name in proj1.c, function descriptions, inline comments).
You should not use any of the functions defined in the math library. The only exception is the use of pow() function in the computations of the final formula (-1)^s * M * 2^E in get_float_from_bits() function.

Deliverables

Implementation if functions in proj1.c file.

Expected output from prob3.c:

Test 0 PASSED. 
    value 2.010000e+02
    bit vecotr 01000011010010010000000000000000 
    2.010000e+02 is normalized 
    E = 7
    M = 1.10010010000000000000000
    s = 0

Test 1 PASSED. 
    value 1.200000e+38
    bit vecotr 01111110101101001000111001010010 
    1.200000e+38 is normalized 
    E = 126
    M = 1.01101001000111001010010
    s = 0

Test 2 PASSED. 
    value -1.024250e+03
    bit vecotr 11000100100000000000100000000000 
    -1.024250e+03 is normalized 
    E = 10
    M = 1.00000000000100000000000
    s = 1

Test 3 PASSED. 
    value 1.401298e-45
    bit vecotr 00000000000000000000000000000001 
    1.401298e-45 is denormalized 
    E = -126
    M = 0.00000000000000000000001
    s = 0

Test 4 PASSED. 
    value inf
    bit vecotr 01111111100000000000000000000000 
      inf is special 
    E = 255
    M =
    s = 0

Test 5 PASSED. 
    value -nan
    bit vecotr 11111111110000000000000000000000 
     -nan is special 
    E = 255
    M = 
    s = 1

Expected output from prob4.c:

Test 0 PASSED. 
              bit vector:  01000011010010010000000000000000
           correct value:  2.010000e+02
          computed value:  2.010000e+02
                           E: 7 
                           M: 1.570312e+00

Test 1 PASSED. 
              bit vector:  01111110101101001000111001010010
           correct value:  1.200000e+38
          computed value:  1.200000e+38
                           E: 126 
                           M: 1.410593e+00

Test 2 PASSED. 
              bit vector:  11000100100000000000100000000000
           correct value:  -1.024250e+03
          computed value:  -1.024250e+03
                           E: 10 
                           M: 1.000244e+00

Test 3 PASSED. 
              bit vector:  00000000000000000000000000000001
           correct value:  1.401298e-45
          computed value:  1.401298e-45
                           E: -126 
                           M: 1.192093e-07

Test 4 PASSED. 
              bit vector:  01111111100000000000000000000000
           correct value:  inf
          computed value:  inf
                           E: 255
                           M: inf

Test 5 PASSED. 
              bit vector:  11111111110000000000000000000000
           correct value:  -nan
          computed value:  -nan
                           E: 255
                           M: nan

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