Answer-first summary for fast verification
Answer: 1024 – 4096
## Explanation This question tests understanding of file system allocation methods and cluster/block sizes. ### Key Points: 1. **File System Allocation**: The system uses chained allocation with an allocation table (likely FAT - File Allocation Table). 2. **Cluster Size**: 4096 bytes (4KB) per block/cluster. 3. **Code Analysis**: - The program writes 1 byte at a time - The loop runs 1024 times (`for(i=0;i<1024;i++)`) - Total data written: 1024 bytes ### Analysis: 1. **File Size (Logical Size)**: The file contains exactly 1024 bytes because that's what was written to it. 2. **Space Occupied (Physical Size)**: In file systems with fixed cluster sizes, space is allocated in whole clusters. Since the cluster size is 4096 bytes: - 1024 bytes requires at least 1 cluster - 1 cluster = 4096 bytes - Therefore, the file occupies 4096 bytes on disk ### Why Other Options Are Incorrect: - **A (1024 – 1024)**: Incorrect because it ignores cluster allocation overhead - **B (1025 – 1025)**: Incorrect - the loop writes 1024 bytes, not 1025 - **D (1025 – 4096)**: Incorrect - the file contains 1024 bytes, not 1025 - **E (4096 – 4096)**: Incorrect - the file contains 1024 bytes, not 4096 ### File System Concept: In chained allocation with allocation tables (like FAT), files are stored in clusters. Even if a file uses only part of a cluster, the entire cluster is allocated to that file. This is known as **internal fragmentation** - wasted space within allocated clusters. **Correct Answer: C (1024 – 4096)**
Author: Danyel Barboza
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A hard disk partition is formatted with a file system based on chained allocation, using an allocation table. The partition formatting considered a block (cluster) size of 4096 bytes. A program executed on this computer creates a file in this partition and writes the file data according to the code snippet below. As a result, the created file contains ____ bytes and occupies ____ bytes in the file system's data area.
...
05 char dado;
...
25 for(i=0;i<1024;i++)
27 write(fd, &dado, 1);
28 close(fd);
...
05 char dado;
...
25 for(i=0;i<1024;i++)
27 write(fd, &dado, 1);
28 close(fd);
Select the alternative that correctly and respectively fills in the blanks in the text above.
A
1024 – 1024
B
1025 – 1025
C
1024 – 4096
D
1025 – 4096
E
4096 – 4096