C程序：当我使用malloc函数时，我发现总是有更多的字节而不是数组的大小

Question

Maybe the new code can explain what I mean.

#include <stdio.h>
#include <stdlib.h>
int main(void){
    int i_size = 2, j_size = 5, k_size = 5, l_size = 6, m_size = 6;
    int i = 0, j = 0, k = 0, l = 0, m = 0; 
    printf("ptr1\n");
    int** ptr1 = (int**) malloc(sizeof(int*) * i_size);
    for(i = 0; i < i_size; i++){
        printf("%d\n", ptr1 + i);
    }
    printf("ptr2\n");
    int** ptr2 =  (int**) malloc(sizeof(int*) * j_size);
    for(j = 0; j < j_size; j++){
        printf("%d\n", ptr2 + j);
    }
    printf("ptr3\n");
    int** ptr3 =  (int**) malloc(sizeof(int*) * k_size);
    for(k = 0; k < k_size; k++){
        printf("%d\n", ptr3 + k);
    }
    printf("ptr4\n");
    int** ptr4 =  (int**) malloc(sizeof(int*) * l_size);
    for(l = 0; l < l_size; l++){
        printf("%d\n", ptr4 + l);
    }
    printf("ptr5\n");
    int** ptr5 =  (int**) malloc(sizeof(int*) * m_size);
    for(m = 0; m < m_size; m++){
         printf("%d\n", ptr5 + m);
    }
    free(ptr1);
    free(ptr2);
    free(ptr3);
    free(ptr4);
    free(ptr5); 
    return 0;
}

The result is as below:

It is funny that it seems for malloc function, the minimun size of array is 32 bytes. And the OS seems to preserve 8 bytes space because when I found that for l_size , it has 64 bytes. It means that if you want to malloc a 48-byte array, the OS will allocate more than 48-byte space.

update above

=====

My code is shown as below:

#include <stdio.h>
#include <stdlib.h>
int main(void){
    int i_size = 5, j_size = 6, i = 0, j = 0; 
    printf("Addresses of two dimensional pointer\n");
    int** ptr = (int**) malloc(sizeof(int*) * i_size);
    printf("%d:\t%d\n", &ptr, ptr);
    for(i = 0; i < i_size; i++){
        *(ptr + i) = (int*) malloc(sizeof(int) * j_size);
    }
    for(i = 0; i < i_size; i++){
        for(j = 0; j < j_size; j++){
            *(*(ptr+i) + j) = i * 2 + j;
        }
    }
    for(i = 0; i < i_size; i++){
        printf("%d:\t%d\n", ptr + i, *(ptr + i));
    }
    printf("==\n");
    for(i = 0; i < i_size; i++){
        for(j = 0; j < j_size; j++){
            printf("%d:\t%d\t%d\n",*(ptr + i) + j, *(*(ptr + i) + j), ptr[i][j]);
        }
        printf("==\n");
    }
    for(i = 0; i < i_size; i++){
        free(*(ptr + i));
    } 
    free(ptr);

    return 0; 
}

And the result as the folloing picture shows:

The code is running on the windows 10, 64-bit and compiled with TDM-GCC4.9.2 of Dev C++.

I am confused by the red block. Why is always 8 more bytes than the size of array. If I change the value of i_size to 6, it seems that the OS will give 64 bytes to ptr but not 48.

I expect that the value 12021664 should be 12021648 . If I change the value of i_size to 7, it is ok:

But when the value of i_size is 8, I expect the value 10710960 should be 10710944 . But it doesn't work.

Answer 1

ptr points to an array of int * . On your system, it appears that pointers are 8 bytes in size, so each member of the array is 8 bytes.

When you then print the address of each member as ptr + i , you see that they each differ by 8.

Also, you should be using %p to print pointers instead of %d . Using the wrong format specifier invokes undefined behavior .

Regarding the specific memory addresses returned by malloc , those are an implementation detail of the library. There is no requirement that successive allocations should be adjacent in memory. In fact, it makes sense that they are not adjacent because there is likely some metadata being stored in those in-between addresses which is being used internally by malloc .