Pointer Arithmetic (Comparing the Pointers)

C Programming: Pointer Arithmetic in C Programming.
Topic discussed:
1) Comparing two pointers using relational and equality operators.
2) The example problem of pointer comparison.
3) Homework problem on comparison of pointers.

Music:
Axol x Alex Skrindo - You [NCS Release]

#CProgrammingByNeso #CProgramming #Pointers #PointerArithmetic #PointersInC

[mohamedmezid]

The correct answer should be option (a) "45 7 4 1 1"

Explanation:

-*(p+3) will give the value at the memory location 4 elements ahead of p, which is a[5] = 45
-*(q-3) will give the value at the memory location 3 elements behind q, which is a[2] = 7
-q-p will give the difference in the number of elements between the two pointers which is (5-1) = 4
-p<q will give the result of the comparison between the memory addresses of the two pointers, is 1 (true) since the address of p is less than the address of q
-*p<*q will give the result of the comparison between the values pointed by the two pointers, which is 1 (true) since the value pointed by p is less than the value pointed by q.
.*p and *q are the values of the pointers which are 16 and 32 respectively.

[hemalathayr]

p+3=45
q-3=7
q-p=1020-1004/4=4
p<q compare addresses
*p<*q compare values
correct answer (A)

[tathagatadas]

for Homework assignment, option 1 is the correct option according to me.

[MAYANKGUDADHECI-]

you are the only one teacher in YouTube community that help student to learn languages in deep all basics is clear

[rahuldorai]

Thank you for all your hard work.You have explained in the bestest way and as short as possible ❤️❤️
Regarding the answer it will be A.

[amirghorban]

*(p+3) = 45
*(q-3)= 7
After this
Now return initial value to variable means *p =16 and *q = 45
q-p = indexof(q) - indexof(p) =4
Adress of p < q ? Yes so return 1
Is 16 < 45 ? Yes return 1

[rajivkumawat]

opyion a is correct.
p+3=45
q-3=7
q-p=1020-1004/4=4
p<q compare addresses
*p<*q compare values
correct answer (A)

[umairalvi]

Just to make things more clear in the above lecture in first example the neso guy in the statement
Printf("%d\n", p<=q); is comparing the address value and not the actual values in the array which are 4 and 6.if you also change the values then also the output will stay the same because the comparison is in between the address values and the address value of a[3] is less than the a[5].

[amanpandey]

Correct Answer is opt :A
If by chance anyone getting 47 7 4 1 0 then it's totally wrong because there's no doubt changing pointing position of pointer but remember we have never stored it in form of p and q and result of last printf is true i.e output will be 1 hence correct answer is opt:A
(ALL are openly invited if I am wrong anyhow)

[olzhasdossaibek]

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[the_ocean_eyes]

1. *(p+3) = *(&a[1+3 ]) = &a[4] = 45
2. *(q-3) = *(&a[5-3]) = &a[2] = 7
Let, address of a[0] = 1000 So,
3. q-p = 1020 - 1004 = 16 = 4*4(bytes) = 4
4. p<q = 1004 < 1020 = 1(true)
5. *p < *q = 7 < 32 = 1(true)

Output - (A) 45 7 4 1 1

[abhimanyunag]

Option A is the correct answer of the homework problem

[vudhanthineeraja]

Haven't used my compiler nor looked at the comment section. I'm confident now. The answer is A.

[zubiarfan]

Because many say option A is correct: It is NOT! Copy the program, save it as a C file and compile it. You will see the output will be: 45 7 4 0 1. So option D is CORRECT!

Why? Here is the answer:

**Answer:**

Initially, `*p` points to 16 and `*q` points to 32.

The following are the various `printf()` statements:

1. `*(p + 3)` = `*(&a[1 + 3])` = `*(&a[4])` = 45
2. `*(q - 3)` = `*(&a[5 - 3])` = `*(&a[2])` = 7
3. `q - p` = 1020-1004/4=4
4. `q < p` = 1020 < 1004 = 0 (`false`)
5. `*p < *q` = 16 < 32 = 1 (`true`)

[Surya-kpqv]

man this stuff is making me get through my first semester, haha. You got some great videos!!

[Subratajana]

We are de-referencing in the first two print statements but we know that de-referencing can never change his address but it can change the value stored in that address so his address will be same as the old addresses. thus right answer is no 1

[Karthik-orfy]

I know most of the people already commented the ans but still I will comment 😂 Ans:a

[satyamgawade]

assuming adresses starting from 100 ie &a[0] is 100

*(p+3)===*(116)===45
*(q-3)===*(108)===7
q-p===(120-104)/4===4
p<q===104 < 120 ===-T===1
*p < *q===16 < 32===T===1

[ganavin]

A is the correct option....it helps to know much about pointers ...tq... keep

[nikhilanand]

sir, so pointer will store the address as initialized earlier but will point to different locations as mentioned.