Chapter 6 solutions Problem [A][c]

Question-

int i = 0 ; 
main( ) 
{ 
printf ( "\nmain's i = %d", i ) ; 
i++ ; 
val( ) ; 
printf ( "\nmain's i = %d", i ) ; 
val( ) ; 
} 
val( ) 
{ 
i = 100 ; 
printf ( "\nval's i = %d", i ) ; 
i++ ; 
} 

Solution-
Easy problem but only confuses sometimes.so, let's see first i is a global variable. so changes made in the value of i would last ,returned accordingly , doesn't matter if you use call by value way.first i=0 ,1st printf statement would print 0. next,i++ would make value of i=1; in the function val(), i would become 100. so second printf statement, which in inside first function,would print 100.next i++ would make i=101. third printf statement would print i=101. it is changed value of i was changes by i++ in first function call. that's really fascinating about global variable.now,second function call would again make i=100 and final printf statement would print i=100. then , again i would be increment by i++ to i=101. if we put one more printf statement in the program, we can verify it.so finally results are
i=0;
i=100;
i=101;
i=100;