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Good question! It’s essential to allocate the correct amount of memory for the data types we want to store.

When you use malloc(n * sizeof(double)), you're telling the program to allocate memory for n number of double values. Each double typically takes 8 bytes of memory. Therefore, by calculating n * sizeof(double), you're ensuring you get enough space for all the doubles you intend to store. If we only allocated ptr with an address without specifying the correct size, there might not be enough memory for all your doubles, leading to errors or program crashes.

So, we need to explicitly allocate n * sizeof(double) to ensure we have enough memory for all n double values you want to work with. Hope this helps!

Hello again, Stephen! I'm not exactly clear about what you're asking, but I'm guessing you're asking one of two questions (or both) I've answered below.

Q1) What's the mathematical operation we're performing?

The calculate_sum() function calculates the sum from 1 to n, where n is the argument passed to it.

Here, n = 3, so calculate_sum() should return:

calculate_sum(3) = 1 + 2 + 3 = 6

==============================================

Q2) How does this recursive function work?

Recursion is a confusing topic that even many of us still struggle with from time to time.

What's important to remember about recursion is this:

The recursive function doesn't start returning values until the base case is reached.

In our case, the base case is n == 1.

So, calculate_sum(3) will recursively call itself until calculate_sum(1) is called.

Once we've reached this base case, calculate_sum(1) will return a value to calculate_sum(2), which in turn returns a value to calculate_sum(3).

Here's the sequence of the recursive calls:

calculate_sum(3) --->calculate_sum(2) --->calculate_sum(1)

Here's the sequence of value returns:

calculate_sum(1) --->calculate_sum(2) --->calculate_sum(3)

I'll try to explain it to the best of my abilities. Please try to read this explanation alongside the image we've provided in our Recursion course:

Part 1: Successive Function Calls

1. First, result = calculate_sum(3) is executed, which calls the calculate_sum() function with 3 as argument. Let's simply call this function call as calculate_sum(3)

2. Notice the statement return n + calculate_sum(n - 1);. Inside calculate_sum(3), this statement becomes return 3 + calculate_sum(2).

3. Here, the return statement isn't fully executed. Instead, it calls calculate_sum(2) first.

4. In other words, a separate instance of calculate_sum() is called (with 2 as an argument). Thus, calculate_sum(2) is separate from calculate_sum(3). Please remember that no value has been returned so far.

5. Similarly, calculate_sum(2) executes return 2 + calculate_sum(1). This means a separate function calculate_sum(1) is called without returning any value.

Part 2: Successive Value Returns

1. We left off at when calculate_sum(1) is called. Now, let's pick off right where we left off.

2. calculate_sum(1) directly returns the value 1 without calling any further functions. That's because calculate_sum(1) is the base case where the if (n == 1) statement is executed.

3. This return value of calculate_sum(1) then goes back to calculate_sum(2).

4. Remember that calculate_sum(2) executes return 2 + calculate_sum(1). Since calculate_sum(1) returned 1, this return statement becomes return 2 + 1, i.e., return 3.

5. In other words, calculate_sum(2) returns the value 3 to calculate_sum(3).

6. Then, calculate_sum(3) executes return 3 + calculate_sum(2), which is return 3 + 3, i.e., return 6.

7. Thus, calculate_sum(3) returns the value 6 to result.

And that's how we get 6 as the final result.

Hope that helps! Contact us again if you have further questions.

Hi Alice 😊

A command line argument is a value you pass to a program when you run it from the terminal. These values tell the program what data it should work with or what action it should perform.

For example, if you have a program that adds numbers, you can run it like this:

./add 10 20

Here, 10 and 20 are the command line arguments. The program reads these values and uses them to calculate the result.

I hope this explains the concept clearly 😊 Please let me know if you need any further help.

Hi there! Great question, let’s clear it up for you.

A pointer is indeed a type of variable in C programming. But, unlike regular variables that store actual values, a pointer stores the address of a variable in memory.

Think of a pointer like a directional sign pointing to where something is stored, rather than what is stored there. When you create a pointer for an integer with int* pt;, you're essentially saying, "I want pt to hold the address of a memory location that contains an integer." This is why we use the symbol * in pointers—it denotes that the variable is meant to store a memory address.

So, yes, while a pointer is a kind of variable, it differentiates itself by holding addresses rather than values directly. Remember, the variable that pt points to will be manipulated by accessing the address stored in pt.

Hope this helps! Let me know if you have more questions or need further clarification.

Hi there!

Your confusion is completely natural because there are a lot of advanced and deeper aspects of C programming that we haven't taught yet (because they're beyond the scope of our course).

Basically, you're asking why we don't need to print \n right after printing employee1.name, right? That's because of how fgets works.

Basically, fgets not only stores the name that you entered inside employee1.name, but it also stores the newline character \n.

How does this happen?

To give the name input, we type the name and then press the Enter key. Hitting the enter key inputs the newline character \n into the input stream, which fgets ends up storing inside employee1.name.

So if you've entered Brad Pitt as input, fgets will store it like this:

employee1.name = "Brad Pitt\n\0"

Note: Please remember that \0 is the null terminator that's present at the end of all strings in C.

So when you print employee1.name, the \n is already included in that variable. That's why you use

printf("%s %d", employee1.name, employee1.age);

instead of

printf("%s\n%d", employee1.name, employee1.age);

But scanf("%s", string_var) behaves differently

On the other hand, if you use something like this:

scanf("%s", employee1.name) 

then the newline character does not get stored inside employee1.name because scanf throws out everything that comes after a whitespace (including the whitespace itself).

The drawback here is that you can only store a single word.

What's the solution?

The solution I gave you before is good enough, but it's still not perfect because employee1.name stores \n at the end.

This can cause a lot of problems in more complex programs because a name string (or any other string) should only contain the required data (i.e., the name) without unnecessary whitespaces.

So the solution is to trim the string after using fgets.

You can remove/trim the string by following the steps below:

1. Import the string.h header file by using #include .

2. Use strcspn() to get the index of the \n character in the string.

3. Then, replace the character at that index with the null terminating character \0 so that \n gets completely removed.

Here's how to do this:

Using strcspn() to get index of \n

// Get the index of \n
int newline_index = strcspn(employee1.name, "\n");

Here, strcspn(employee1.name, "\n") compares two strings: employee1.name and "\n", and gives the index of the first occurrence of the second string i.e., "\n".

Since employee1.name has only one instance of \n in it, strcspn() effectively gives us the position of the extra line at the end of this string.

Then, we remove this newline character by replacing it with \0:

// Replace the \n with \0
employee1.name[newline_index] = '\0';

You can also simply combine the two steps like this:

employee1.name[strcspn(employee1.name, "\n")] = '\0';

Here's the FULL SOLUTION:

#include 
#include 

// create Person struct
struct Employee {
  int age;
  char name[50];
};

int main() {

  // create struct variable
  struct Employee employee1;

  // get age input for employee1's age
  scanf("%d", &employee1.age);
  getchar();

  // get name input for employee1's name
  fgets(employee1.name, sizeof(employee1.name), stdin);

  // Get the index of \n
  int newline_index = strcspn(employee1.name, "\n");

  // Replace the \n with \0
  employee1.name[newline_index] = '\0';
  
  // print name and age
  printf("%s\n", employee1.name);
  printf("%d", employee1.age);

  return 0;
}

Note: It's better to use employee1.name[strcspn(employee1.name, "\n")] = '\0'; instead of storing the \n index in a separate variable. The whole point of C programming is to be memory-efficient.

Why is this not included in the course (or in my previous answers)?

I wanted to inform you about this issue yesterday but thought the answer might get too long and confusing.

So I'm thankful that you asked this followup question.

Currently, our course cannot go into such deep detail because it's aimed at absolute beginners.

For now, I can only tell learners how to properly deal with it when they ask questions, as you've done here.

I hope you found this helpful! Please don't hesitate to ask further questions if you're still confused.

Hi again! You don't need null terminators for an array of numbers. You only need to add null terminators at the end of strings.

This is because strings in C are just character arrays. But how do you know if a char array is a string or just an ordinary group of characters?

For example, suppose you have the following arrays:

// not a C string, just a simple char array
char vowels[5] = { 'a', 'e', 'i', 'o', 'u' };

// is a C string because of \0
char greeting1[6] = { 'H', 'e', 'l', 'l', 'o', '\0' };

// also a C string (compiler adds '\0')
char greeting2[] = "Hello";

Basically, normal char arrays don't have \0 at the end, but strings do. The presence of \0 is how C knows whether something is a string or not.

So, you don't need to put \0 in arrays except for strings.

Hope that helps!

When you use fptr == NULL, you're actually checking if fptr is equal to the special value NULL, which indicates that no file was opened successfully at that pointer.

Basically, fopen() is designed to either:

• return a pointer to a FILE object representing the file you want to work with,

• or, if there's an error (like if the file isn't found), it returns NULL.

• The value NULL in this context signals that the pointer fptr doesn't point to any valid file, allowing you to safely check if the file operation succeeded.

In other words, NULL isn't just an ordinary value; it's actually a special marker that's treated differently from ordinary literals such as 1, 0, "yes", "no", etc.

That's why you can indeed check fptr == NULL but can't do checks such as fptr == "yes" since that's not allowed in pointer operations.

Hope this helps you understand the behavior of NULL and pointers better! Let me know if you have any more questions.

Hi again! A short answer to this particular question is included in my answer to your previous question.

If you're still confused, you can try the following program and see what happens:

#include 

int main() {

  int age;

  // get age input
  printf("Enter age: ");
  scanf("%d\n", &age);
  
  // print age
  printf("Age: %d", age);

  return 0;
}

You'll see that this program "hangs" after you've entered the age and pressed enter, i.e., it doesn't print the age after you've given the input.

The only way to make it print the age is to give another input and press enter. Here's my full output:

Enter age: 25
a
Age: 25

Notice that I had to enter a (or any other non-whitespace input) to print the age.

Obviously, this is bad because the program appears to be hanging/freezing in the middle of the execution.

And as I've explained in my previous answer, this happens because you're telling scanf() to consume all whitespace characters and wait for non-whitespace character input.

That's what scanf("%d\n", &variable); does, and it's not something you want in your code.

Hope this helps! Let me know if you have any more questions.

Hi there! At first glance, your code appears to make intuitive sense. But thanks to the way C programming functions internally, you've ended up with several problems in your code.

The two major issues are:

1. Using \n with scanf

Here's how you've taken input for age:

scanf("%d\n", &enployee1.age);

Please don't do this. In scanf format strings, whitespace characters (including \n) mean: consume any amount of whitespace, then keep waiting until the next non-whitespace character appears.

So scanf("%d\n", ...) will:

• read the integer,

• then consume the newline you typed,

• and then block until you type the first non-whitespace character of the next input (for example, the first letter of the name).

To fix this, please remove \n from this code like this:

scanf("%d", &enployee1.age);

Then consume the leftover newline before reading a line of text:

getchar(); 

Here, getchar() will consume the newline character \n entered after the age input so it doesn't interfere with the next input statement.

2. Getting individual character input for "name" and printing the individual characters instead of printing the name as a single string.

On the surface, it makes sense to use a loop to get individual character inputs for a string (since scanf() can only take input for a single word).

But there are many problems with this:

• Using a loop from 0 to 49 means the code will always take input for 50 characters, even if the actual name contains far less characters (say, 20 characters). This is wasteful.

• This loop also doesn't add a null terminating character \0 at the end of the string.

• Printing each character of the string is not ideal. It's better to print the entire name as a string rather than individual characters.

To fix this, please use fgets() to get string input like this:

fgets(employee1.name, sizeof(employee1.name), stdin);

A Minor Spelling Mistake

You've misspelled Employee by writing it as Enployee, and your object names also have this mistake.

This doesn't matter within the code but it's still better to use proper spellings.

Full Solution

#include 

// create Person struct
struct Employee {
  int age;
  char name[50];
};

int main() {

  // create struct variable
  struct Employee employee1;

  // get age input for employee1's age
  scanf("%d", &employee1.age);

  // use getchar() to consume the newline
  getchar();

  // get name input for employee1's name
  fgets(employee1.name, sizeof(employee1.name), stdin);

  // print name and age
  printf("%s", employee1.name);
  printf("%d", employee1.age);

  return 0;
}

Hello there! It's natural to be confused about the sizeof operator in C, and whether it includes the null terminator as well.

To answer your question: no, we don't need to write sizeof(name) + 1 because sizeof measures the full size of the variable (in bytes), which includes the null terminator as well.

Hope that helps! Contact us again if you have further questions.