**Educational Text on C Declaration and Memory Layout**---Consider the following C declaration and assume the following for the machine:- **1-byte characters**- **2-byte shorts**- **4-byte integers**- **4-byte single-precision floating-point numbers**- **4-byte pointers**- **8-byte double-precision floating-point numbers**The compiler does not reorder fields and leaves no memory gaps. All other data types are considered as 8-byte:```cunion U1 { int value; char code; double data; void* pointer;};union U2 { float func; short small[2];};union U3 { char text; long long number;};union U4 { char name[6]; unsigned int id;};struct S { int index; union U1 item; union U4 person; void* next;} s;```**Questions to Consider:**1. **How many bytes does each `union` declared in the code and `struct S` occupy?**2. **If the memory address of `S` starts from 3000, what are the start addresses of `S.item.pointer` and `S.person.name[2]`?**3. **If the compiler uses a simple memory alignment mechanism that aligns variables to 8-byte boundaries, what will be the size of `S`?**---**Explanation and Analysis:**1. **Union Sizes:** - Each union is sized according to its largest member. - `union U1`: Largest member is `double data` (8 bytes). - `union U2`: Largest member is `func` or `small[2]` (4 bytes each, totaling 8 bytes). - `union U3`: Largest member is `number` (8 bytes). - `union U4`: Largest member is `id` (4 bytes, but in practice unions are aligned to larger sizes).2. **Struct S Size:** - Includes size of `index` (4 bytes), `union U1` (8 bytes), `union U4` (aligned to 8 bytes), and `next` (4 bytes). - Alignment adjustments may lead to additional padding.3. **Address Calculations:** - Given starting address 3000, calculate addresses based on sizes and alignments.This exercise emphasizes understanding memory

Name: **Educational Text on C Declaration and Memory Layout**---Consider th
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Description: **Educational Text on C Declaration and Memory Layout**---Consider the following C declaration and assume the following for the machine:- **1-byte characters**- **2-byte shorts**- **4-byte integers**- **4-byte single-precision floating-point numbers

In this question we have to understand about data structures, memory layout, and alignment in the…

Engineering

Computer Science

Consider the following C declaration and assume that the machine has 1-byte characters, 2-byte short, 4-byte integers, 4-byte single-precision floating point numbers, 4-byte pointers, 8-byte double- precision floating-point numbers, and the compiler does not reorder fields and leaves no memory gaps. You may consider all other data types as 8-byte. union UI ( int value; char code; double data; void pointer; }; union U2 ( float func; short small [2]; }; union U3 ( B; char text; long long number; }; union U4 ( char name [6]; unsigned int id; }; struct S ( int index; union Ui item; union U4 person; void next;

Consider the following C declaration and assume that the machine has 1-byte characters, 2-byte short, 4-byte integers, 4-byte single-precision floating point numbers, 4-byte pointers, 8-byte double- precision floating-point numbers, and the compiler does not reorder fields and leaves no memory gaps. You may consider all other data types as 8-byte. union UI ( int value; char code; double data; void pointer; }; union U2 ( float func; short small [2]; }; union U3 ( B; char text; long long number; }; union U4 ( char name [6]; unsigned int id; }; struct S ( int index; union Ui item; union U4 person; void next;

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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