#define min((a),(b)) ((a)<(b))?(a):(b) main() { int i=0,a[20],*p

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Question

#define min((a),(b))  ((a)<(b))?(a):(b)
	main()
	{
	 int i=0,a[20],*ptr;
	 ptr=a;
	 while(min(ptr++,&a[9])<&a[8]) i=i+1;
	 printf("i=%d\n",i);}

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Re: #define min((a),(b)) ((a)<(b))?(a):(b) main() { int i=0,a[20],*ptr; ptr=a; while(min(ptr++,&a[9])<&a[8]) i=i+1; printf("i=%d\n",i);}

Answer
# 1

Is This Answer Correct ?

2 Yes

2 No

Guest

Re: #define min((a),(b)) ((a)<(b))?(a):(b) main() { int i=0,a[20],*ptr; ptr=a; while(min(ptr++,&a[9])<&a[8]) i=i+1; printf("i=%d\n",i);}

Answer
# 2

here the value will be 3.

EXPLANATION:
here the #define macros will blindly substitute the values
in the while loop before compailation.. then when it compails...
1)we will have the expanded macros like this:
            while((ptr++<&a[9]?ptr++:&a[9])<&a[8])
i++;

hrer when the loop runs for the first time ptr will
increment by 2 since it is a integer type, which allocates 2
bytes.
first see the layout:
 10  12  14    16    18   20   22    24   26   28  30 ..... 
|   |   |    |     |     |    |     |    |    |   |   |    
  0   1    2     3    4     5   6     7    8    9   10 ....
0,1,2,3 represents index values...
10,12,14 represents addresses....
so the ptr variable will have the base address of array a.
when comin to while loop, it gets incremented to the next
location , address is 12,and go and check wit the address of
&a[9] in our case it is 28. so naturally it wil become true
so it executes the statement after ? symbol.. again in that
ptr++ is given so again ptr will be incremented to 14.. so
14 will be compared with &a[8] ,likely to be 26. it is TRUE
so the whole loop is true ,so i gets incremented so i=1.
next time ptr adds to 16 and then 16<28 and again ptr gets
incremented to 18 and 18<26 and whole while becomes true so
i will become 2.
similarly it will again increments ptr to 20 an dthis
becomes true an ptr again gets incremented to 22 and it
checks whether 22<26! yes, it will increment i to 3.

IMPORTANT:
after that it increments ptr to 24 and executes the
operation after ? operator and again ptr will have 26, this
26 is checked with 26 (&a[8]) the whole while loop becomes
false...
so it wont go to i++, it will printf the printf statement so
i=3.

Is This Answer Correct ?

2 Yes

1 No

Vignesh1988i

Re: #define min((a),(b)) ((a)<(b))?(a):(b) main() { int i=0,a[20],*ptr; ptr=a; while(min(ptr++,&a[9])<&a[8]) i=i+1; printf("i=%d\n",i);}

Answer
# 3

Is This Answer Correct ?

0 Yes

0 No

Question	Asked @	Answers
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