for (j = 2; j <= (i / j); j++) if (!(i % j)) break; if (j > (i / j)) printf("%d ", i); return 0;
Understand the Problem
The question provides a snippet of code and seems to ask about its behavior, potentially related to what output it would generate or how it works. The focus is on understanding the logic of the code provided.
Answer
The code checks for prime numbers and prints them.
The final answer is the code checks if 'i' is a prime number and prints it if true.
Answer for screen readers
The final answer is the code checks if 'i' is a prime number and prints it if true.
More Information
This code snippet completes a common task in programming: finding and displaying prime numbers using a nested loop approach.
Tips
A common mistake is not understanding the condition 'j <= (i / j)' which optimizes checking till the square root of 'i'.
Sources
- This is a program to find prime numbers from 2 to 100 in C [closed] - stackoverflow.com
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