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C Structs

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Dynamically Allocating Structs

Let's review how to dynamically allocate memory in the C language.

To allocate memory on the heap, use the malloc() function from the stdlib.h library:

<pointer> = (data_type*)malloc(n * sizeof(data_type));

Note

When you dynamically allocate memory for use in your program, that memory remains reserved until you explicitly free it using free().

main.c

main.c

copy

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

// structure definition
struct Example {
    int someValue[20]; // int = 4 bytes, 20 * 4 = 80 bytes
};

int main() {
    // allocating memory for Example structure
    struct Example* pExample = (struct Example*)malloc(sizeof(struct Example));

    printf("Allocating memory = %zu bytes\n", sizeof(*pExample));

    // freeing memory
    free(pExample);
    
    return 0;
}

If a structure contains several fields of different data types, the compiler will equalize the size of the fields so that the structure is "conveniently" stored in memory.

After the completed job, pay attention to how many bytes are allocated for the structure with the int and char fields. This phenomenon will be discussed later in this course.

Uppgift

Swipe to start coding

  1. Create a pointer variable and allocate memory for your structure;
  2. Display the size of structure;
  3. Fill the fields and display the content of your structure;
  4. Free up the allocated memory;
  5. Displaying the values again.

Lösning

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book
Dynamically Allocating Structs

Let's review how to dynamically allocate memory in the C language.

To allocate memory on the heap, use the malloc() function from the stdlib.h library:

<pointer> = (data_type*)malloc(n * sizeof(data_type));

Note

When you dynamically allocate memory for use in your program, that memory remains reserved until you explicitly free it using free().

main.c

main.c

copy

              
12345678910111213141516171819

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

// structure definition
struct Example {
    int someValue[20]; // int = 4 bytes, 20 * 4 = 80 bytes
};

int main() {
    // allocating memory for Example structure
    struct Example* pExample = (struct Example*)malloc(sizeof(struct Example));

    printf("Allocating memory = %zu bytes\n", sizeof(*pExample));

    // freeing memory
    free(pExample);
    
    return 0;
}

If a structure contains several fields of different data types, the compiler will equalize the size of the fields so that the structure is "conveniently" stored in memory.

After the completed job, pay attention to how many bytes are allocated for the structure with the int and char fields. This phenomenon will be discussed later in this course.

Uppgift

Swipe to start coding

  1. Create a pointer variable and allocate memory for your structure;
  2. Display the size of structure;
  3. Fill the fields and display the content of your structure;
  4. Free up the allocated memory;
  5. Displaying the values again.

Lösning

Switch to desktopByt till skrivbordet för praktisk övningFortsätt där du är med ett av alternativen nedan
Var allt tydligt?

Hur kan vi förbättra det?

Tack för dina kommentarer!

close

Awesome!

Completion rate improved to 4.17

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