What will be the output of the following code:

#include <stdio.h>

void main()
{
int i;

for (i=0;i<2;i  )
{
    fork();

    printf("1\n");
}

printf("2\n");

}

On the lectures, the professor ran the code and the output had six 1s and four 2s, but I don't understand how we obtained those numbers.

CodePudding user response：

Lets "draw" the processes in a tree form, and mark out where the printf call are happening, and then count the printf calls of each type:

                      parent
                        |
                      fork
                        ^
         parent -------/ \------------ child
           |                             |
        printf("1")                   printf("1")
           |                             |
     loop iterates                 loop iterates
           |                             |
          fork                          fork
           ^                             ^
 parent --/ \----- grandchild   child --/ \----- grandchild
   |                  |           |                 |
printf("1")       printf("1")  printf("1")       printf("1")
   |                  |           |                 |
loop ends         loop ends    loop ends         loop ends
   |                  |           |                 |
printf("2")       printf("2")  printf("2")       printf("2")
   |                  |           |                 |
proc ends         proc ends    proc ends         proc ends

As we can see there are six printf("1") calls, and four printf("2") calls.

_{[I haven't added the newlines in the printf calls, the diagram only is for counting the number of calls. Actual output may differ depending on buffering depending on stdout target.]}