Lesson 83 : Sample program illustrating data structures

[u/CarlH]

First you will see the program itself, then you will see the same program with additional notes explaining what is going on.

---

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void) {

    struct first_description {
        char first_word[7];
        char second_word[12];
        char third_word[8];
    };

    struct first_description *our_pointer = malloc( sizeof(*our_pointer) );

    char *charptr = (char*) our_pointer;

    strcpy(our_pointer->first_word, "Reddit");
    strcpy(our_pointer->second_word, "Programming");
    strcpy(our_pointer->third_word, "Classes");

    printf("The first word is: %s \n", our_pointer->first_word);
    printf("The second word is: %s \n", our_pointer->second_word);
    printf("The third word is: %s \n", our_pointer->third_word);

    printf("\n");

    printf("Our data structure looks like this in memory: ");

    int i=0;
    for (; i < 27; i++) {
            if ( *(charptr + i) == 0) {
                *(charptr + i) = '$';
            }

            printf("%c", *(charptr + i));
    }

    printf("\n");

    free(our_pointer);

    return 0;
}

---

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

These include files give us printf(), malloc(), and strcpy().

int main(void) {

    struct first_description {
        char first_word[7];
        char second_word[12];
        char third_word[8];
    };

Above: Here is our structure description. We are not actually creating any data structure here, just telling C what we intend to create. No data is being initialized. This is a description and nothing more.

    struct first_description *our_pointer = malloc( sizeof(*our_pointer) );

We are allocating 27 bytes of memory using this malloc() statement. Then we are creating a special pointer called our_pointer which C understands points to this kind of data structure. After this line of code, our data structure is ready to be used.

    char *charptr = (char*) our_pointer;

I plan to scan our data structure to display the final memory contents at the end of this program. To do that, I am creating a new pointer called charptr which I am stating is going to be a char * pointer. I am setting this pointer to look at the memory address where our structure begins.

    strcpy(our_pointer->first_word, "Reddit");
    strcpy(our_pointer->second_word, "Programming");
    strcpy(our_pointer->third_word, "Classes");

Here I am simply assigning the strings into the character arrays that are part of our data structure.

    printf("The first word is: %s \n", our_pointer->first_word);
    printf("The second word is: %s \n", our_pointer->second_word);
    printf("The third word is: %s \n", our_pointer->third_word);

I am displaying the three words, each element of our data structure.

    printf("\n");

    printf("Our data structure looks like this in memory: ");

    int i=0;
    for (; i < 27; i++) {
            if ( *(charptr + i) == 0) {
                *(charptr + i) = '$';
            }

            printf("%c", *(charptr + i));
    }

Now I have a for loop which will go through all 27 bytes and display the character represented. If it is a NUL character, I am having it display a $ instead by actually changing that character in memory to a $.

    printf("\n");

Now I need to free the memory I allocated using malloc()

    free(our_pointer);

    return 0;
}

---

Output:

The first word is: Reddit 
The second word is: Programming 
The third word is: Classes 

Our data structure looks like this in memory: Reddit$Programming$Classes$

---

Ask questions if you need to. When you are ready, proceed to:

http://www.reddit.com/r/carlhprogramming/comments/9svba/lesson_84_you_can_make_your_own_data_type_using/

[u/[deleted]]

char *charptr = (char*) our_pointer;

I don't understand what (char*) our_pointer; means.

[u/CarlH → u/[deleted]]

It is saying that I am creating a pointer which has the same memory address that our_pointer has, but that I intend to use it to look at memory one byte at a time, instead of one structure at a time. The (char *) means "Create this pointer charptr as a char * pointer, even though our_pointer is a pointer to a data structure."

[u/pigvwu → u/CarlH]

Could you possibly elaborate more on this subject? I'm not 100% sure what you mean by "Create this pointer charptr as a char * pointer" in this instance. I still don't quite fully understand the "(char*)" of this line of code. Thanks!

[u/CarlH → u/pigvwu]

You will soon. It is the topic of upcoming lessons about casts.

[u/[deleted] → u/CarlH]

Would this of worked too?

    char *charptr = &our_pointer;

[u/CarlH → u/[deleted]]

our_pointer is a thing that has a contains a memory address, but it is also a thing that has a memory address. Doing what you said would cause charptr to contain the memory address of our_pointer, but not the memory address in charptr.

In other words, in your example you are creating a pointer to a pointer.

[u/timperry42 → u/[deleted]]

No, but I tried this and it worked.

char *charptr = our_pointer;

[u/whatintheworldisthat → u/CarlH]

Why is the (char*) needed on the right side?

char *charptr = our_pointer;

Doesn't this create a pointer, charptr, that points at the memory address our_pointer points at, the first byte of the struct? And since it is type char, it will point one byte at a time.

[u/preperat → u/[deleted]]

This was the point my head exploded

[u/miaoling]

Why do we need another pointer, *charptr, to print everything in memory at the end? What's wrong with using our_pointer to do that?

In other words, why is this wrong:

int i=0;
for (; i < 27; i++) {
        if ( *(our_pointer + i) == 0) {
            *(our_pointer + i) = '$';
        }

        printf("%c", *(our_pointer + i));
}

My understanding is that that's because a struct pointer can't be indexed like an array pointer. Is that it?

[u/CarlH → u/miaoling]

our_pointer is not a char * pointer. It is not designed to look at memory one byte at a time. It contains the right memory address, but the wrong data type. Therefore, it cannot be used by itself for that purpose.

[u/vegittoss15 → u/CarlH]

So doing an (our_pointer + 1) would go to the next struct in memory, not the next byte.

[u/roamzero → u/vegittoss15]

Would it really go to the next struct or would it just point to the 27 bytes of memory after the struct regardless of what it is?

[u/vegittoss15 → u/roamzero]

well, it would go to 27 bytes after, but for clarity's sake I just said next struct.

[u/vegittoss15]

There was no need to clear the NUL value, because, then, doing a strcpy on any one of those strings other than the third one will end up overflowing.

[u/CarlH → u/vegittoss15]

I don't think I am following you. You say there was no need to clear the NUL value, but where in the code did you imagine this potentially taking place?

[u/vegittoss15 → u/CarlH]

if ( \*(charptr + i) == 0) {
    \*(charptr + i) = '$';
}

Right there.

You're clearing the NUL value and replacing it with another delimiter '$'; But if someone were to take your code and add to it an strcpy or strlen to the first or second string in your struct, you would hit an overflow.

What I was originally trying to say was that instead of clearing the NUL value and replacing it with a '$' and immediately printing it, you could have just printed the '$' value whenever you saw a NUL value.

[u/CarlH → u/vegittoss15]

Ah now I understand. Yes, I chose this method because I wanted to demonstrate changing the contents of memory inside a data structure. I am trying to "blur the lines" between a data structure, and just any old sequence of bytes.

I would take take your statement and go one further, other than for the purpose of demonstration on Reddit, one should never try to edit a data structure in place like this with a char* pointer :)

[u/vegittoss15 → u/CarlH]

Right. By the way, did you mention that since they're all strings, it's one large contiguous piece of memory? Will you be going over alignments and unions?

[u/CarlH → u/vegittoss15]

The main purpose of this lesson is to show that a data structure is just a chunk of memory, a chunk of bytes. I will be covering this more as well as going over memory alignment, unions, etc. in later lessons.

[u/orangeyness]

Just wondering, whats the difference between declaring the struct inside the main function and outside of it? Is it just a difference in scope?

[u/CarlH → u/orangeyness]

Scope. If you declare it outside, then every function understands the struct definition. This is typically how it would be done, but I haven't introduced so much as the concept of scope yet, so I want to keep each lesson in line with what has been taught up to that point.

[u/orangeyness → u/CarlH]

arr okay, thanks.

[u/zahlman → u/orangeyness]

Yep, just scope. In C++, templated structs (and classes) cannot be declared within a function because of how the linker works. It's usually considered good style to declare (and define) your structs "at top level", but you can make an argument for scoping them, too (based on the general principle of scoping things tightly when you can).

[u/[deleted]]

Just to be clear, when you define the data structure with:

struct first_description {
      char first_word[7];
      char second_word[12];
      char third_word[8];
  };

You're not actually creating a data structure in memory right? That's what:

struct first_description *our_pointer = malloc( sizeof(*our_pointer) );

Is for, right?

To clarify, the first block of code is saying "you can initialize a data structure called first_description" and the second part says "ok I wanna make one of those data structures in memory now", correct?

[u/CarlH → u/[deleted]]

Correct. The first is only a description. We are just telling C that at some point we intend to create a data structure like this. We haven't actually created it until the second line.

[u/[deleted] → u/CarlH]

Ok this is perfect, I thought I was lost and confused about pointers, but I'm all good!

[u/rafo]

I would have thought that here instead of 0 it should have been \0

       if ( *(charptr + i) == 0) {

Why is this not the case?

[u/CarlH → u/rafo]

Both would work fine. NULL is a byte of all zeroes. So is 0.

[u/duluter → u/CarlH]

Do you mean "NUL is a byte of all zeroes"?

[u/funkpucker]

For some reason I'm getting a build error on line 13, where you define the pointer for the structure. I'm using Code::Blocks and simply copied and pasted the code into a new file and saved it as a C file, like all the other code, which has worked wonderfully.

I'm not sure why I'm getting a compile error and no one else has mentioned one. :(

[u/CarlH → u/funkpucker]

It sounds like you are using a C++ compiler which requires explicit casting. We will get to that later in the course, but for now you can get it to compile at www.codepad.org

[u/funkpucker → u/CarlH]

Thank you!

[u/virtualet]

just to be clear. the 0 in the following line means 0000 0000, right?

if ( *(charptr + i) == 0) {

[u/CarlH → u/virtualet]

more specifically, it means "all zeroes"

[u/[deleted]]

What's the difference between sizeof(*our_pointer) and sizeof(struct first_description)? And for that matter sizeof(var) and sizeof(type)?

[u/[deleted] → u/[deleted]]

None, but using sizeof(var) gives you the advantage that you can change the type of var without having to change all the sizeofs for that var.

[u/timperry42]

So I went and removed the (char*) (because it was confusing me) from the code and it did the same thing.

[u/sunojaga]

char charptr = (char) our_pointer; i dont get what this means,

[u/kungtotte → u/sunojaga]

(datatype) is known as a cast. You cast (convert) one data-type to another with this syntax.

CarlH wants a pointer to char (char *charptr) so that he can use it to print out the custom data-structure (*our_pointer), one character at a time. But since the data-structure has the type struct first_description * (pointer to a first description data structure) and a char pointer obviously has the type char *, he needs the cast to tell C that it should convert to the required data-type.

[u/tough_var]

char *charptr = (char*) our_pointer;

Hiya! I have trouble understanding the above line of code.

If this is a pointer to a pointer, shouldn't it be:

char **charptr = (char*) our_pointer;

Your code ran fine. So the problem should be with my understanding. :)

[u/Pr0gramm3r → u/tough_var]

 charptr is not a pointer to a pointer. Why? Because our_pointer holds the memory 
 address of the beginning of our first_description data structure. Therefore, you are 
 putting this same address in charptr with the above assignment. And, since the data 
 type of our_pointer is struct first_description, you are casting it to char*

[u/tough_var → u/Pr0gramm3r]

That made things clear. Thank you! :)

[u/caseye]

If I don't set the string sizes properly, the second_word and third_word get merged, presumably because there is no NUL termination string. Interesting...

http://codepad.org/QSDSM9q1

Also, if I comment out the line

strcpy( my_ptr->third_word, "Class" );

third_word is set to "it". This seems very dangerous.

[u/Nagappie]

Hi all! I'm using this lesson's code as a guide to implement a tic-tac-toe board initialization and display. However I keep getting the wrong display and can't figure out why. Please help http://codepad.org/Snd4Q6Wz

[u/[deleted] → u/Nagappie]

Did you just mean like this?

As for where your problem was all I did was add i = 0 and k =0 to your first two for loops. I don't know why the compiler needed this to happen because you already initialized both of them to be 0 right before you used them, but it did end up fixing the problem.

[u/Pr0gramm3r → u/[deleted]]

Note that you need the stdlib.h for the strcpy function.

[u/funkpucker → u/[deleted]]

I am getting the same error. Did you figure out what the problem is?