How to calculate sizeof() in c programming manually (without using computer)?? H

Question

How to calculate sizeof() in c programming manually (without using computer)??

Here are some examples, I don't know how these answers (sizeof) were calculated??

(5 points) What's the output of this code snippet? (%zu prints a value of unsigned long integer, %p prints the address of a pointer) int (*p)[10]; p malloc(sizeofin) * 20); assume malloc() is successful printf"'sizeofp) %zum", sizeofp)); printf"sizeof*p)-%zuln", sizeof(*p)); printf("pl-%pm", (void *)p++); printf"p2 %pin", (void *)p); Output: sizeof(p) (1 point) sizeof(*p) - 40 (2 points)

Explanation / Answer

#define my_sizeof(type) (char *)(&type+1)-(char*)(&type)

int main()

{

    double x;

    printf("%d", my_sizeof(x));

    getchar();

    return 0;

}

######################################################################

int main()

{

   int arr[] = {1, 2, 3, 4, 7, 98, 0, 12, 35, 99, 14};

   printf("Number of elements :%d", sizeof(arr)/sizeof(arr[0]));

   return 0;

}

############################################################################

int *ptr = malloc(10*sizeof(int));

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