Question

Run the following code. Discuss the output result, i.e. the value of seq variable.

#include <stdio.h>

#include <unistd.h>

int main()

{ int seq = 0; if(fork()==0) { printf("Child! Seq=%d\n", ++seq); }

else { printf("Parent! Seq=%d\n", ++seq);

} printf("Both! Seq=%d\n", ++seq); return 0; }

Answer 1

Answer

Given code:

#include<stdio.h>
#include<unistd.h>
int main()
{
int seq = 0;
if(fork()==0) {
printf("Child! Seq=%d\n", ++seq);
}
else
{
printf("Parent! Seq=%d\n", ++seq);
}
printf("Both! Seq=%d\n", ++seq); return 0;
}

Output:

Explanation:

• When the main function executes, seq is initialized to zero & if statement is false because there is no fork() (Child Process) is executed so else part will be printed that is
  "parent! seq =1" where seq is pre incremented.
• Then the statement if Main() function will be executed that's "Both! seq =2", where previous value of seq was increamented to 2 from 1.
• In the Execution of if statement fork() creates Child Process, which starts from Main() as:
• a- Seq initializes to 0.
• b- 'If' statment becomes true because fork() was created in Main() program execution. So "if" part will be printed that's "Child! seq=1".
• c- Similarly statment of main() will be printed as "Both! seq =2", where seq is incremented to 2 from 1.

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