**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 context of C programming.It covers several concepts:Union and Struct Sizes: Assessing knowledge of how the size of unions and structures is determined based on the sizes of their members.Memory Addresses: Testing ability to calculate the memory addresses of specific members within a structure.Memory Alignment: Evaluating understanding of how memory alignment affects the size of structures and the placement of their members in memory.Let's understand and hope this helps, If you have any queries please utilize threaded question feature.Size of Unions:Union U1:int (4 bytes) + char (1 byte) + double (8 bytes) + void* (4 bytes) = 17 bytesUnion U2:float (4 bytes) + short[2] (2 shorts, each 2 bytes) = 8 bytesUnion U3:char* (4 bytes) + long long (8 bytes) = 12 bytesUnion U4:char[6] (6 bytes) + unsigned int (4 bytes) = 10 bytesSize of Struct S:int (4 bytes) + Union U1 (17 bytes) + Union U4 (10 bytes) + void* (4 bytes) = 35 bytesSo, the size of each union is as follows:Union U1: 17 bytesUnion U2: 8 bytesUnion U3: 12 bytesUnion U4: 10 bytesThe size of the struct S is 35 bytes.To calculate the start addresses of S.item.pointer and S.person.name[2], we need to consider the sizes of the preceding members in the struct S.Given that the memory address of S starts from 3000 and the size of S is 35 bytes.So, we can calculate the start addresses as follows:Start address of S.item.pointer:S.index (4 bytes) has a starting address of 3000.S.item follows S.index. Since S.item is a union, its size is the maximum of the sizes of its members, which is Union U1 with a size of 17 bytes.Therefore, the start address of S.item is 3000 + 4 = 3004.S.item.pointer is the first member of Union U1, so its start address is the same as the start address of Union U1, which is 3004.Start address of S.person.name[2]:S.index (4 bytes) has a starting address of 3000.S.item follows S.index, and we already calculated its size as 17 bytes, so the start address of S.item is 3004.S.person follows S.item. The start address of S.person is 3004 + 17 = 3021.S.person.name is an array of characters, and the start address of an array is the same as the start address of the array variable. Therefore, the start address of S.person.name is 3021.S.person.name[2] is the third element of the array, so its start address is 3021 + 2 = 3023.Therefore, the start address of S.item.pointer is 3004, and the start address of S.person.name[2] is 3023.3. If the compiler uses a simple memory alignment mechanism that aligns variables to 8-byte boundaries, what will be the size of S?With a memory alignment mechanism that aligns variables to 8-byte boundaries, the size of the struct S will be adjusted to ensure proper alignment of its members. Original Size of Struct S:int (4 bytes) + Union U1 (17 bytes) + Union U4 (10 bytes) + void* (4 bytes) = 35 bytesAdjusted Size for 8-Byte Alignment:To align the struct members properly, the compiler will add padding after the int member to ensure that the start of Union U1 is aligned on an 8-byte boundary. The adjusted size will include this padding.Padding after int (4 bytes) to align Union U1 = 4 bytesAdjusted Size:int (4 bytes) + Padding (4 bytes) + Union U1 (17 bytes) + Union U4 (10 bytes) + void* (4 bytes) = 39 bytesSo, with an 8-byte alignment mechanism, the adjusted size of the struct S will be 39 bytes.

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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