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Lab 1: C

Deadline: Tuesday, June 25, 11:59:59 PM PT

Before you come to lab 1, make sure that you are comfortable with either editing your files on the hive by using your text editor of choice. This lab is designed to familiarize you with basic C concepts and prepare you for project 1.

We expect your submission to follow the spirit of the question. If your submission does not follow the spirit of the question (e.g. hard-codes the correct output while not applying any of the concepts introduced in this lab), we may manually grade your submission. If you feel like you may potentially be violating the spirit of the question, feel free to make a private question on Ed or ask your lab TA.

Lab Slides


Setup

You must complete this lab on the hive machines. See Lab 0 for a refresher on using them.

In your labs directory, pull the files for this lab with:

git pull starter main

If you get an error like the following:

fatal: 'starter' does not appear to be a git repository
fatal: Could not read from remote repository.

make sure to set the starter remote as follows:

git remote add starter https://github.com/61c-teach/su24-lab-starter.git

and run the original command again.

If you run into any git errors, please check out the common errors page.


Exercise 1: Hello!

Compiling and Running a C Program

In this lab, we will be using the command line program gcc to compile programs in C. Use the following command to compile the code for exercise 1 (make sure that you have cded into the proper directory)

gcc ex1_hello.c

This compiles ex1_hello.c into an executable file named a.out. If you've taken CS61B or have experience with Java, you can kinda think of gcc as the C equivalent of javac. This file can be run with the following command:

./a.out

The executable file is a.out, so what is the ./ for? Answer: when you want to execute an executable, you need to prepend the path to the name of the executable. The dot refers to the "current directory." Double dots (..) would refer to the directory one level up.

gcc has various command line options which you are encouraged to explore. In this lab, however, we will only be using -o, which is used to specify the name of the executable file that gcc creates. By default, the name of the executable generated by gcc is a.out. You can use the following commands to compile ex1_hello.c into a program named ex1_hello, and then run it. This is helpful if you don't want all of your executable files to be named a.out.

gcc -o ex1_hello ex1_hello.c
./ex1_hello

At this point, you should see the string Hello World printed out. If you edit the source code (such as ex1_hello.c), you must recompile the program using gcc to produce a new executable, otherwise the executable will still run the old source code.

Edit ex1_hello.c with your editor of choice and make the program print out the string "Hello 61C" instead of "Hello World". Make sure to save your edited file, but do NOT recompile yet.

Run the executable with ./ex1_hello. You should still see Hello World, which is the old output.

Now, recompile your program with gcc -o ex1_hello ex1_hello.c, and then run the executable with ./ex1_hello again. You should now see Hello 61C.

Review: Pointers

A pointer is a variable whose value is the memory address of another variable. Note that every variable declaration is always located in a memory, where every element has a corresponding address. Think of it like an array: every variable value is contained on a specific array index (address), and the pointer to that variable is another variable within that same array that contains the index (address) of the variable it is pointing at.

Consider the following example:

int main() {
    int my_var = 20;
    int* my_var_p;
    my_var_p = &my_var;
}

For the first line, we declared an int variable called my_var which is then assigned with a value of 20. That value of 20 will be placed somewhere in the memory.

For the second line, we declared an int pointer variable called my_var_p. Note that you can also write int *my_var_p, where the asterisk glued to the variable name instead of the variable type.

For the third line, we assigned my_var_p to have a value that is equal to the address of my_var. This is done by using the & operator before the my_var variable. At this point, the value contained in the variable my_var_p is the address in memory of the variable my_var.

Note that whenever you want to change the value of my_var, you could do it by changing my_var directly.

my_var += 2;

Alternatively, you could also change the value of my_var by dereferencing my_var_p

*my_var_p += 2;

In a nutshell, &x gets the address of x, while *x gets the contents at address x.

For this section, assume that sizeof(int) == 4. Here's a more complete example:

int main() {
    int my_var = 20;
    int* my_var_p;
    my_var_p = &my_var;

    printf("Address of my_var: %p\n", my_var_p);
    printf("Address of my_var: %p\n", &my_var);
    printf("Address of my_var_p: %p\n", &my_var_p);

    *my_var_p += 2;

    printf("my_var: %d\n", my_var);
    printf("my_var: %d\n", *my_var_p);
}

A sample execution of this code gave out the following:

Address of my_var: 0x7fffebafb32c
Address of my_var: 0x7fffebafb32c
Address pf my_var_p: 0x7fffebafb330
my_var: 22
my_var: 22

The first line prints out the value of my_var_p, which was assigned to the address of the variable my_var.

The second line shows that my_var_p is indeed equal to &my_var, the address of the variable my_var.

The third line prints out the address of my_var_p. Note that since my_var_p is in fact a variable itself (the variable type is an int pointer), therefore it has to be placed somewhere in the memory as well. Thus, printing out &my_var_p allows us to see where in the memory the my_var_p variable is located.

After the first three print outputs, we changed the value of my_var indirectly using *my_var_p. Since my_var_p is a pointer to my_var (i.e. my_var_p is the address of my_var), performing *my_var_p allows us to modify the contents at the address in my_var_p.

The fourth line shows that we have indeed modified my_var, since the value is now 22.

The fifth line confirms that *my_var_p is indeed equal to my_var.

What happens if we did the following: my_var_p += 2?

my_var_p is a pointer to an int. sizeof(int) = 4. Therefore, pointer arithmetic will add 2 * 4 = 8 The value in my_var_p updates to 0x7fffebafb32c + 8 = 0x7fffebafb334

After doing my_var_p += 2 earlier, what is the value of &my_var_p?

The address of my_var_p will remain the same: 0x7fffebafb330

After doing my_var_p += 2 earlier, what happens then if we try to print the value of *my_var_p?

Since the value of my_var_p has changed, it is now pointing to a different location in memory. The behavior is undefined. It will either print out a garbage data on that specific memory location, or the program crashes with a segmentation fault if you try to access a protected memory segment.

Exercise 2: Pointers and Functions

Edit ex2_pointers_and_functions.c using your editor of choice and fill in the blanks. Feel free to refer back to the pointer review section if you are stuck.

Compile and run the program and check that the output matches what you expect. If you need a refresher on gcc, please refer back to exercise 1.

Exercise 3: Pointers to Stack vs Heap

Edit ex3_ptr_heap_stack.c using your editor of choice and fill in the blanks. Feel free to refer back to the pointer review section if you are stuck.

Compile and run the program and check that the output matches what you expect. You will see a compiler warning. What does it mean?

Read the output of gcc. Note that it throws an error about the address of a stack variable being returned. Similarly, the output of the program should show address of the stack variable as (nil), indicating that it is not usable. This is because x cannot be accessed outside of the function int_on_stack using a pointer. In the future, make sure to allocate memory on the heap if you'd like to use it later.

Review: Arrays

An array is a fixed length data structure that can hold one or more elements of the same type. Unlike lists, arrays do not resize automatically when you add an element.

In C, arrays are represented as a pointer to the first element. Each element of the array is stored in memory and they are stored in contiguous memory locations (side by side). Because arrays are represented only with a pointer to the first element, the pointer itself is not enough to deduce the length of the array. If you need to keep track of the length of an array, you must use another variable.

You can perform arithmetics on pointers to access different elements of the array. Recall that a pointer is just an address, so if you add to or subtract from the address, you can get the address of later or earlier elements of an array.

Exercise 4: Arrays

Edit ex4_arrays.c using your editor of choice and fill in the blanks. Feel free to refer back to the arrays section if you are stuck.

Compile and run the program and check that the output matches what you expect.

Read the output of your program. Note that the relationship between the address of the start of the array and the address of index 2 (hint: they're two bytes apart).

Review: Pointer Arithmetic

In exercise 4, your program performed basic pointer arithmetic when you computed the address of index 2. This worked because the size of each element is 1 byte (since the size of int8_t is 1 byte). However, most types you'll work with take up more than 1 byte in memory.

When performing pointer arithmetic, C automatically accounts for the type of the pointer and adds the correct number of bytes. For example, if you write ptr + 5, C will not always add 5 to ptr. Instead, C will add 5 times the size of the datatype that ptr points to. If ptr was an int* and ints take up 4 bytes in memory, ptr + 5 adds 20 to the address held in ptr.

Exercise 5: Pointer Arithmetic

Edit ex5_pointer_arithmetic.c using your editor of choice and fill in the blanks. Feel free to refer back to the arrays review section if you are stuck. Your solution should be similar to exercise 4.

Compile and run the program and check that the output matches what you expect.

Read the output of your program. Note that the relationship between the address of the start of the array and the address of index 2 is different from the relationship in the previous exercise.

Review: Strings

In C, strings are represented as an array of chars. Strings are a special type of char arrays because they always end in a null terminator (\0). Recall that arrays in C do not contain any information about their length, so the null terminator allows us to determine when the string ends.

When allocating memory for a string, there must be enough memory to store the characters within the string and the null terminator. Otherwise, you might run into undefined behavior. However, the array could be larger than the string it stores.

C has a library of functions for manipulating strings, such as:

  • strlen: computes the length of a string by counting the number of characters before a null terminator
  • strcpy: copies a string from one memory location to another, one character at a time until it reaches a null terminator (the null terminator is copied as well)

Exercise 6: Strings

Edit ex6_strings.c using your editor of choice and fill in the blanks. Feel free to refer back to the strings review section if you are stuck.

Compile and run the program and check that the output matches what you expect.

Exercise 7: Copying Strings

Edit ex7_strcpy.c using your editor of choice and fill in the blanks. When copying the string, use strcpy (documentation).

Compile and run the program. Notice that the program crashes, why?

Why does it crash?

strcpy is an unsafe function. We only allocate enough space in message to store the message hello, but then we try to store a longer message. When this happens, we overflow the space we allocated for message, and crash the program!

Fix the program in ex7_strcpy_fixed.c using strncpy (documentation) so that it stores as many characters of longer_message as you can in message without changing the size of message.

Compile and run the program again and check that the output matches what you expect.

Exercise 8: Structs

Edit ex8_structs.c using your editor of choice and fill in the blanks.

Compile and run the program and check that the output matches what you expect.

Optional: typedefs

Sometimes, you may see a typedef when declaring a struct:

typedef struct {
    int id;
} Student;

In these cases, you may use Student as the type instead of struct Student. We won't go into detail here, but feel free to check out this link if you're interested.

Exercise 9: Reflection and Feedback Form

We are working to improve the class every week - please fill out this survey to tell us about your experience in CS 61C so far!


Submission

Save, commit, and push your work, then submit to the Lab 1 assignment on Gradescope.


FAQ

What is a header file?

Header files allow you to share functions and macros across different source files. For more info, see the GCC header docs.

What is a null terminator?

A null terminator is a character used to denote the end of a string in C. The null terminator is written as '\0'. The ASCII value of the null terminator is 0. When you make a character array, you should terminate the array with a null terminator like this

char my_str[] = {'e', 'x', 'a', 'm', 'p', 'l', 'e', '\0'};

If you are using double quotes to create a string, the null terminator is implicitly added, so you should not add it yourself. For example:

char *my_str = "example";

What is an executable?

An executable is a file composed of binary that can be executed on your computer. Executables are created by compiling source code.

What is strlen?

See the man pages for a full description. Type the following into your terminal

man strlen

To exit the man pages, press q.

What is a macro?

A macro is a chunk of text that has a name. Whenever this name appears in code, the preprocessor replaces the name with the text. Macros are indicated with #define For example:

#define ARR_SIZE 1024
#define min(X, Y)  ((X) < (Y) ? (X) : (Y))

int main() {
    int arr1[ARR_SIZE];
    int arr2[ARR_SIZE];
    int arr3[ARR_SIZE];

    for (int i = 0; i < ARR_SIZE; ++i) {
        arr3[i] = min(arr1[i], arr2[i]);
    }
}

In this code, the preprocessor will replace ARR_SIZE with 1024, and it will replace

arr3[i] = min(arr1[i], arr2[i]);

with

arr3[i] = ((arr1[i]) < (arr2[i]) ? (arr1[i]) : (arr2[i]));

Macros can be much more complex than the example above. You can find more information in the GCC docs

What is a segfault?

A segfault occurs when you try to access a piece of memory that "does not belong to you." There are several things that can cause a segfault including

  1. Accessing an array out of bounds. Note that accessing an array out of bounds will not always lead to a segfault. The index at which a segfault will occur is somewhat unpredictable.
  2. Derefrencing a null pointer.
  3. Accessing a pointer that has been free'd (free is not in the scope of this lab).
  4. Attempting to write to read-only memory. For example, strings created with the following syntax are read only. This means that you cannot alter the value of the string after you have created it. In other words, it is immutable.
char *my_str = "Hello";

However, a string created using the following syntax is mutable.

char my_str[] = "hello";

Why is the first string immutable while the second string is mutable? The first string is stored in the data portion of memory which is read-only while the second string is stored on the stack.