I want to know why structure variable passes pointer variable to create memory.

1. what happens if we do box *boxes = malloc(n * sizeof(box));

2. Then we pass address to scanf function. Pointer actually stores the address. Then why we pass "&" of scanf to pointer ?

for (int i = 0; i < n; i  ) {
    scanf("%d%d%d", &boxes[i].length, &boxes[i].width, &boxes[i].height);
}

#include <stdio.h>
#include <stdlib.h>
#define MAX_HEIGHT 41

struct box
{
    /**
    * Define three fields of type int: length, width and height
    */
    int length,width,height;
};

typedef struct box box;

int get_volume(box b) {
    /**
    * Return the volume of the box
    */
    return b.length*b.width*b.height;
}

int is_lower_than_max_height(box b) {
    /**
    * Return 1 if the box's height is lower than MAX_HEIGHT and 0 otherwise
    */
    return b.height < 41 ? 1 : 0;
}

int main()
{
    int n;
    scanf("%d", &n);
    box *boxes = malloc(n * sizeof(box));
    for (int i = 0; i < n; i  ) {
        scanf("%d%d%d", &boxes[i].length, &boxes[i].width, &boxes[i].height);
    }
    for (int i = 0; i < n; i  ) {
        if (is_lower_than_max_height(boxes[i])) {
            printf("%d\n", get_volume(boxes[i]));
        }
    }
    return 0;
}

• When I tried running the sizeof struct,I got the sizeof Struct box it was 12.
• Say n = 5, Then what will be the memory space? memory = 12 * 5 ?

CodePudding user response：

1. malloc() is used to allocate dynamic and variable sized memory. So we use this to create an array of n structures.

2. You need & because scanf() needs the address of the variable to store the input data to. b is a pointer, but b[i].length is just an ordinary structure member accessed by dereferencing that pointer. You need to get its address to pass to scanf().

CodePudding user response：

 box *boxes = malloc(n * sizeof(box));

Allocates enough space for n instances of box on the heap and returns a pointer to it

 scanf("%d%d%d", &boxes[i].length///

is passing the address of boxes[i].length to scanf. Scanf needs to write there so its needs the address. Its just like

scanf("%d", &n);

that you did earlier

 sizeof(struct box)

equaling 12 is fine. It has 3, 4 byte integers. You wont always get it that simply , the compiler might pad the structure to align things on 2, 4 or 8 byte boundaries, this is why sizeof exists

Say n = 5, Then what will be the memory space? memory = 12 * 5 ?

yes