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iable is stored in memory, it is associated with an address. To obtain the address of a variable, the & operator can be used. Write a C program addressOfScalar.c by inserting the code below in the main function. Questions: 1) Run the C program, attach a screenshot of the output in the answer sheet. 2) Attach the source co

iable is stored in memory, it is associated with an address. To obtain the address of a variable, the & operator can be used. Write a C program addressOfScalar.c by inserting the code below in the main function. Questions: 1) Run the C program, attach a screenshot of the output in the answer sheet. 2) Attach the source co

Database System Concepts
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
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Part 1: When a variable is stored in memory, it is associated with an address. To obtain the address of a variable, the & operator can be used.  Write a C program addressOfScalar.c by inserting the code below in the main function. Questions:

 

 

1) Run the C program, attach a screenshot of the output in the answer sheet.

2) Attach the source code in the answer sheet

3) Then explain why the address after intvar is incremented by 4 bytes instead of 1 byte.

```c 1 // initialize a char variable, print its address and the next address 2 char charvar = '\0'; 3 printf("address of charvar = %p\n", (void *)(&charvar)); 4 printf("address of charvar - 1 = %p\n", (void *)(&charvar - 1)); 5 printf("address of charvar + 1 = %p\n", (void *)(&charvar + 1)); 6 7 // initialize an int variable, print its address and the next address 8 int intvar = 1; 9 printf("address of intvar = %p\n", (void *)(&intvar)); 10 printf("address of intvar - 1 = %p\n", (void *)(&intvar - 1)); 11 printf("address of intvar + 1 = %p\n", (void *)(&intvar + 1)); 12 ``` ### Explanation The code snippet is written in C and demonstrates how memory addresses are handled for different data types, specifically a `char` and an `int`. 1. **Character Memory Addressing:** - A `char` variable `charvar` is initialized with the null character (`'\0'`). - The program prints the address of `charvar`, then the address one byte before, and one byte after it. - As `char` typically occupies 1 byte, adding or subtracting 1 from its address moves the address by 1 byte. 2. **Integer Memory Addressing:** - An `int` variable `intvar` is initialized with the value `1`. - The program prints the address of `intvar`, then the address one integer size before, and one integer size after it. - As `int` typically occupies 4 bytes (depending on the system architecture), adding or subtracting 1 from its address moves the address by the size of `int` bytes. This code helps understand how memory allocation and pointer arithmetic work in C programming by demonstrating how different data types affect pointer arithmetic due to their distinct sizes in memory.
Transcribed Image Text:```c 1 // initialize a char variable, print its address and the next address 2 char charvar = '\0'; 3 printf("address of charvar = %p\n", (void *)(&charvar)); 4 printf("address of charvar - 1 = %p\n", (void *)(&charvar - 1)); 5 printf("address of charvar + 1 = %p\n", (void *)(&charvar + 1)); 6 7 // initialize an int variable, print its address and the next address 8 int intvar = 1; 9 printf("address of intvar = %p\n", (void *)(&intvar)); 10 printf("address of intvar - 1 = %p\n", (void *)(&intvar - 1)); 11 printf("address of intvar + 1 = %p\n", (void *)(&intvar + 1)); 12 ``` ### Explanation The code snippet is written in C and demonstrates how memory addresses are handled for different data types, specifically a `char` and an `int`. 1. **Character Memory Addressing:** - A `char` variable `charvar` is initialized with the null character (`'\0'`). - The program prints the address of `charvar`, then the address one byte before, and one byte after it. - As `char` typically occupies 1 byte, adding or subtracting 1 from its address moves the address by 1 byte. 2. **Integer Memory Addressing:** - An `int` variable `intvar` is initialized with the value `1`. - The program prints the address of `intvar`, then the address one integer size before, and one integer size after it. - As `int` typically occupies 4 bytes (depending on the system architecture), adding or subtracting 1 from its address moves the address by the size of `int` bytes. This code helps understand how memory allocation and pointer arithmetic work in C programming by demonstrating how different data types affect pointer arithmetic due to their distinct sizes in memory.
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