Problem 4 [4pt] Consider the following C declaration: struct Si (char a; int b;

Question

Problem 4 [4pt] Consider the following C declaration: struct Si (char a; int b; ) struct S2 (float a; double b; union U struct Si sl; struct S2 s2; Assume the machine has 1-byte characters, 4-byte integers, 8-byte floating numbers, and 16-byte double- precision floating numbers. Assume the compiler does not reorder the fields, and it leaves no holes in the memory layout. How many bytes does u occupy? If the memory address of u starts from 1000, what are the start addresses of u·s1.b and u . s2.b?

Explanation / Answer

size of struct S1 is = 1(char) + 4(int) = 5 (add 3 bytes for padding) = 8 bytes;

size of struct S2 is = 8 (float) + 16(double ) = 24 bytes (here no padding is required)

/* i'll explain padding at the last */

so the union size is the greater of two so size of union is 24 bytes . As the base address is 1000 the union is occupied upto 1024 bytes.

In union you can only access one variable at a time.

While accessing s1 only 8 bytes are used so

s1.a base address is : 1000

s1.b base address is : 1004

while accessing s2 all the 24 bytes are used so

s2.a base address is : 1000

s2.b base address is : 1008

Padding :

padding is that the compiler makes the size of struct and unions to be multiples 4 so for struct S1 it is made 8 by adding 3 bytes.

/* hope this helps */

/* if any queries please comment */

/* thank you */

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.