Why Does `fork()` Output 8 Dots Instead of 6 in This Code?
Nov 03, 2024 pm 05:32 PM
Fork()'s Unexpected Branching Behavior
Consider the following code that utilizes the fork() system call:
<code class="c">#include <stdio.h>
#include <sys/types.h>
#include <unistd.h>
int main(void)
{
int i;
for(i = 0; i < 2; i++)
{
fork();
printf(".");
}
return 0;
}</code>
Surprisingly, when this program is executed, it outputs 8 dots instead of the expected 6. This discrepancy initially seems puzzling.
Dissecting the Code's Execution
To understand this behavior, we must break down the execution process step by step:
- Initial State: There is a single process executing.
- First Iteration: The process forks, creating a child process with a separate memory space. Both the parent and child print a dot.
- Second Iteration: Both the parent and child fork again, creating a total of four processes. Each prints a dot.
- Process Termination: All four processes terminate, but the buffered dots from earlier print operations remain in the output buffer.
Buffered Output
The crux of the issue lies in the buffered output of printf(). When a process prints, the output is not immediately flushed to the screen. Instead, it is stored in a buffer. This behavior is significant because the fork() system call copies the buffer, resulting in duplicate buffered dots.
A Total of 8 Dots
When the four processes terminate, their buffered dots are flushed, adding them to the single dot printed initially. This results in a total of 8 dots being output.
Avoiding the Issue
To prevent this behavior, one can explicitly flush the output buffer using fflush(stdout) after each printf() call. This ensures that output is immediately sent to the screen, eliminating the discrepancy between the number of processes and the number of printed dots.
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