The code at 14:30 works because the character array C is stored in a contiguous block of memory, so when we increased the pointer C, we were traversed through the character array C by one byte at a time, which is the amount of memory required to store a character. Once the pointer reaches the null terminator, the loop ends.
Good to note that it also works with an array of integers since the incrementation automatically increments it by 4 bytes, which I find to be extremely cool.
Hi HashDLS, Size of pointer variable will always be same whether its a pointer to int or pointer to char. Its an address right. So, It's about how many bytes we want to use to store an address. Now, this could be address of anything. In 2 bytes or 16 bits we can store max 2^16 addresses, in 4 bytes or 32 bits - 2^32. It depends upon compiler and architecture of machine. Turbo -C may be compiling for 16 bit architecture. CodeBlocks would be compiling for 32 bit architecture.
14:30 works because: Since the array is passed by reference to the function, the pointer will point to the base of the array which is index 0. Then because C is passed as a char pointer, incrementing C by one will ensure traversing the array char after char since a char is 1 byte.
Sir you did really great job to demonstrated all concepts in very lucid way, you are rally adept,veteran teacher who delivered such complex topics in such a way to ease way to understandable
Yash Pandey hey guys... the value is'e' only.... because in c1 the values to"hello" will be given as h=0,e=1,l=2,l=3,o=4... So when c2[1 ] is given it gives 'e'.. and as *c2=c1......c2 will be same as character string c1.. So c2=c1... hope u understand...... reply me if u have any kind of doubts regarding that...
For the question at the end, I think it works because the function print(char* C) took a pointer as argument. Well, pointers can be incremented, so starting at the initial address of the string allows us to locate all the characters because they are stored contiguously in memory. Is this right? Thank you.
I dont usually write stuff like this but your video has helped explain alot that ive had trouble with at uni. Wish I had learnt this earlier but better late then never, thanks heaps and great video
*My takeaways:* 1. How to store strings 0:59 2. Arrays and pointers are different types that are used in a similar manner 7:38 3. Arrays are always passed to function by reference 11:20
@@lordstark5292 That´s wrong as it would output the addresses of each individual character in the string. This is because you used the address of operator instead of the dereference operator.
When passing an array to a function, a pointer variable is created by compiler that points at the address of the first element of the array, thus, c (the parameter in print function) is converted to a pointer which points to the address of 'H'. So as pointers can be incremented (to point to the next location) AND array elements are stored contiguous in memory, the code works :)
I think your videos are awesome. But in this video, I want to point one tiny thing, I don't think it matters, but print(char *C), it is passing arguments by value(not reference). In the book P.209 "Pointers behave like any other nonreference type. When we copy a pointer, the value of the pointer is copied. After the copy, the two pointers are distinct. However, a pointer also gives us indirect access to the object to which that pointer points. We can change the value of that object by assigning through the pointer" After I read this, I realized C/C++ has so many things need to pay attention to.
one thing to remenber is when using *(C+i) it is actually doing *(C+i*1) where 1 is size of char in bytes. In case of working with other type of arrays like int(4 byte) double(8 byte) we need to multiply explicitly
as far as i know, when you increment an pointer by 1 it jumps to the next element in the array, if an int is 4 bytes , the pointer jumps 4 bytes to the next integer by incrementing the pointer by 1. maybe you are using some language that doesn't?
@@mihai5085 It does and that's what the comment is explaining. That when you do *(C + i) the compiler actually reads it as *( C + i * sizeof() ) which is how it jumps to the next element instead of the next byte.
At a point in this video he stated that declaring and intiliazing ( char A[4] = "JOHN" ) will give a compilation error but that doesnt happen. I checked it on two IDE,s GEEKSFORGEEKS and DEVC++
Thanks for the video it clear my concepts , there is some mistake you had said forward slash at 3:34 but you had actually used backward slash so please correct it as the other user does not get wrong undrstnding
It is worth it to watch in 2024 ??? , ima beginner in c , i even don't have idea does any code or way of typing of code is changed . Should I follow your code ? Will it work ? For mine
the last exercise could be done this way while(*C) putchar(*C++); putchar(' '); if you dereference of the char pointer it will return the current character that is not zero so the loop will continue until it finds the null terminator.
Yes, it’s possible it may work. Hence the name undefined. But if you keep doing it, you will eventually run into problems. You can drive on the wrong side of the road and be fine, keep doing it and you’re going to run into some big issues.
at 14:30 if I print the value of C after the print function called, does the memory address changed as well? (because we increment it inside the loop in print function)
At 3:32 that's a backslash not a forwardslash. You can remember ti's a backslash because imagine a person walking across the screen.left to right. but he is falling backwards. Generally right is forward and left is backward.. Look where the top of the backslash is, it's back . It is like the slash that Windows command shell uses, CD BACKSLASH CD \, as opposed to the forward slash that linux command shell uses.
A character array doesn't require null termination but if a character array needs to define a string then we add '\0' there to say that it's null terminated and now it is a string, so we can have char array without null termination but then don't expect to use it in strlen() or other functions which expect a string. Peace
In the last example I believe it will also work if the while loop was written : while (*C) . Meaning while *C is true , and it can only be true if it’s not null, is that correct? I suppose we can put it to the test.
When printing c2[1], why is that printing out a letter? If c2 is a pointer to the address of the first character of c1, wouldn't: print c2[1]; print out 201 (the address of the second element of the array)? I could see how running: print *c2[1]; would give you 'e', but that's not what's written.
Hi! After declaring C1 as character array and C2 as pointer to character, is C2 = C1 somehow the same as C2 = &C1[0] and the compiler does know that now C2 is usable as an array? (since we havent declared it with [] anywhere). Thanks!
The code at the end will work...coz when u wrote their *c...u derefrenced a value..named c...in the loop...u printed that derefrenced value....as for the fact that arrays are stored as a reference value in another function and only stores the address of the first variable....now after printing the first variable by drefrencing the address of it ...u incremented it...and in loop u passed the condition not equals to null...so it will terminate their and will give us hello as output...
If you increment the c pointer as in the last example. Do you need rewind it back to the original position, else it forever points at the newly incremented address?
![[LIVE] : ONE ลุมพินี 91 | คู่เอก "คมอาวุธ vs อเล็กเซย์"](http://i.ytimg.com/vi/V0RkEfYFsgM/mqdefault.jpg)
The code at 14:30 works because the character array C is stored in a contiguous block of memory, so when we increased the pointer C, we were traversed through the character array C by one byte at a time, which is the amount of memory required to store a character. Once the pointer reaches the null terminator, the loop ends.
@@akashsahu933 did you even read the rest of the explanation?
@@akashsahu933 SHUT UP!
Good to note that it also works with an array of integers since the incrementation automatically increments it by 4 bytes, which I find to be extremely cool.
This guy explains really well. I have a full book on pointers, but I prefer to watch his videos. Awesome C course.
can i know which book it is? also, is it any good?
@@palashbehra9303 The book is "Understanding pointers in C" by Yashavant Kanetkar. I bought it in India while on a business trip there.
@@keen2461 thanks man
@@palashbehra9303 I used Let Us C and it is pretty good too
This whole channel is gold. Thanks. I'm watching this as I read the K&R C book.
Hi HashDLS,
Size of pointer variable will always be same whether its a pointer to int or pointer to char. Its an address right. So, It's about how many bytes we want to use to store an address. Now, this could be address of anything. In 2 bytes or 16 bits we can store max 2^16 addresses, in 4 bytes or 32 bits - 2^32. It depends upon compiler and architecture of machine. Turbo -C may be compiling for 16 bit architecture. CodeBlocks would be compiling for 32 bit architecture.
Hi!
14:30 works because: Since the array is passed by reference to the function, the pointer will point to the base of the array which is index 0. Then because C is passed as a char pointer, incrementing C by one will ensure traversing the array char after char since a char is 1 byte.
I see everybody is benefiting from your teaching, and I will thank you for doing so. Great simple concise and complete explanation. Thank you
Sir you did really great job to demonstrated all concepts in very lucid way, you are rally adept,veteran teacher who delivered such complex topics in such a way to ease way to understandable
at 9:31 you said that print C2[1] will give us value l ???? But shouldn't it give us the value e ?
+Yash Pandey Yes, we should get e. Honest mistake
Yash Pandey hey guys... the value is'e' only.... because in c1 the values to"hello" will be given as h=0,e=1,l=2,l=3,o=4... So when c2[1 ] is given it gives 'e'.. and as *c2=c1......c2 will be same as character string c1.. So c2=c1... hope u understand...... reply me if u have any kind of doubts regarding that...
sai srikar lol😁 extra marks for steps 👍
it is a mistake.
I like the time difference between the replies , I love how knowledge unites people from not only different places , but different time :)
For the question at the end, I think it works because the function print(char* C) took a pointer as argument. Well, pointers can be incremented, so starting at the initial address of the string allows us to locate all the characters because they are stored contiguously in memory. Is this right? Thank you.
exactly
i am thinking the same
Yupp, you're correct.
I dont usually write stuff like this but your video has helped explain alot that ive had trouble with at uni. Wish I had learnt this earlier but better late then never, thanks heaps and great video
This video didn't become the top viewed in this subject for nothing . Thank you sir
Your videos saved my exams at university
Thank you from Greece!
Well it saved you not your exams lol
You have made a superb video - many thanks.
this channel is just amazing, up there with the best cs channels in the world
awesome work sir, really appreciating, thank you 👍👍
are Deepak sir
Seriously helpful for me to understand this "array" thingy, thanks dude for the video
@mycodeschool ur tutorials are simply the best. thx a lot
9:24 why c2[1] is "l"? maybe it should be "e", as the second element of "hello"
indeed...that seems to be a mistake
Yep you are right that was a mistake
yes it is "e", not "l"
typing error.. you have much concentration.nice
*My takeaways:*
1. How to store strings 0:59
2. Arrays and pointers are different types that are used in a similar manner 7:38
3. Arrays are always passed to function by reference 11:20
Awesome! God bless you. Good example when you first didn't insert and then inserted null character at the end of string.
you are such a great teacher, thanks a lot! your video helped me hugely!
Thank you very much admin you taught really classical
You can also implement print() function like this:
void print(char *s)
{
while (*s)
printf("%c", *s++);
printf("
");
}
Please explain this print function.. as am not able to understand this.
Or like this:
void print(char *c){
putchar(*c++);
if(*c) print(c);
}
@@shivarajpatil162 or like this void print(char *c)
{
int i = 0;
while(c[i])
{
write(1, &c[i], 1);
i++;
}
wow very concise, well done
@@lordstark5292 That´s wrong as it would output the addresses of each individual character in the string. This is because you used the address of operator instead of the dereference operator.
These vidz are fantastic thanks for your time creating and uploading them...
Many many many thanks .... You was very helpfull for me ... Your way in explaining things is more than excellant .... Thank you
OO man , this man is doing magic even after 10 years
Thanks for the `while` idea, solved a problem I was suffering from.
Thanks for your tutorial! You're the best!
Great video's series on pointers. You make it very clear.
thank you!
8:49
Then it can be *c1+i* also(executed inside a for-loop of i)
where, i is the index of the alphabets of the string starting from 0
thanks it helped me in passing character array as argument
When passing an array to a function, a pointer variable is created by compiler that points at the address of the first element of the array, thus, c (the parameter in print function) is converted to a pointer which points to the address of 'H'. So as pointers can be incremented (to point to the next location) AND array elements are stored contiguous in memory, the code works :)
brilliant !
Very easy and efficient explanations.
Thank you so much sir.
u really worked hard. thank u for uploading. it really helped me a lot.
i can use pointer like a expert.
thank you for the time you take to make these amazing videos!
these videos are great sir.. i am recommending this channel to my friends and teacher..
Very beautiful handwriting ... Nice explanation too
Excellent videos in general, the best I've seen so far.
I believe that in a list the null terminated character is implicitly declared ?
I think your videos are awesome. But in this video, I want to point one tiny thing, I don't think it matters, but print(char *C), it is passing arguments by value(not reference). In the book P.209 "Pointers behave like any other nonreference type. When we copy a pointer, the value of the pointer is copied. After the copy, the two pointers are distinct. However, a pointer also gives us indirect access to the object to which that pointer points. We can change the value of that object by assigning through the pointer" After I read this, I realized C/C++ has so many things need to pay attention to.
one thing to remenber is when using *(C+i) it is actually doing *(C+i*1) where 1 is size of char in bytes. In case of working with other type of arrays like int(4 byte) double(8 byte) we need to multiply explicitly
as far as i know, when you increment an pointer by 1 it jumps to the next element in the array, if an int is 4 bytes , the pointer jumps 4 bytes to the next integer by incrementing the pointer by 1. maybe you are using some language that doesn't?
@@mihai5085 you are correct
@@mihai5085 It does and that's what the comment is explaining. That when you do *(C + i) the compiler actually reads it as *( C + i * sizeof() ) which is how it jumps to the next element instead of the next byte.
Amazing Video which is mastered by your hard work
Kudos!!!!
thank you for the great lesson . i searched these information from you tubel a lot.
this may help me for passing this part
very nice tutorial.
thanks a lot.
Really Awesome, great and very clear understanding of the concepts! Thank you very much!
At a point in this video he stated that declaring and intiliazing ( char A[4] = "JOHN" ) will give a compilation error but that doesnt happen. I checked it on two IDE,s GEEKSFORGEEKS and DEVC++
same here i dont know why? have you found out yet?
Thanks for the video it clear my concepts , there is some mistake you had said forward slash at 3:34 but you had actually used backward slash so please correct it as the other user does not get wrong undrstnding
Thank you for clearing my doubts
Amazing video!
Excellent video with the best explanation I have seen yet, thank you!! Plus leaving us homework :)
Wonderful. You’re so awesome.
these videos are gold
Increementing of C works because. in print function it is reference value which can be incremented
thanks for your explaination!
Good quality and depth.
perfect explanation
It is worth it to watch in 2024 ??? , ima beginner in c , i even don't have idea does any code or way of typing of code is changed . Should I follow your code ? Will it work ? For mine
Awesome video for beginner like me ...
your videos are amazing
the last exercise could be done this way
while(*C)
putchar(*C++);
putchar('
');
if you dereference of the char pointer it will return the current character that is not zero so the loop will continue until it finds the null terminator.
u r amazing sir!!
no he is not amazing......he made a lot of mistakes
@@sanambaloch2975 like to share those mistakes?
thankyou so much, your videos are so convincing :-)
Thank you very much! I finally find my bug after watching it.
Didn't use \0 and it still worked.
Yes, it’s possible it may work. Hence the name undefined. But if you keep doing it, you will eventually run into problems. You can drive on the wrong side of the road and be fine, keep doing it and you’re going to run into some big issues.
at 14:30 if I print the value of C after the print function called, does the memory address changed as well? (because we increment it inside the loop in print function)
And. We loved this explanation!👌
very good video .You explain it beautifully . Other series kb ay gi ???
At 3:32 that's a backslash not a forwardslash. You can remember ti's a backslash because imagine a person walking across the screen.left to right. but he is falling backwards. Generally right is forward and left is backward.. Look where the top of the backslash is, it's back . It is like the slash that Windows command shell uses, CD BACKSLASH CD \, as opposed to the forward slash that linux command shell uses.
C++ will work, because you are incrementing the start address that was given as a parameter!
This is gold.
A character array doesn't require null termination but if a character array needs to define a string then we add '\0' there to say that it's null terminated and now it is a string, so we can have char array without null termination but then don't expect to use it in strlen() or other functions which expect a string. Peace
Who's here in 2019?
Where did he go ???
@@baderbboukheit3604 he´s in the other side.
@@adolfocarrillo248 it will be 2020
here in 2020
@Señor Mike no he didn't
I Like Your Voice..! :p
his voice is calm and it makes you feel he is very confident :)
Because C is a pointer to a character. And C++ means to increment the value of C by 1 byte (the address)
at 3:33 its backward slash
Many thanks
In the last example I believe it will also work if the while loop was written : while (*C) . Meaning while *C is true , and it can only be true if it’s not null, is that correct? I suppose we can put it to the test.
amazing explaining, thank you :)
thank u very much you are simply amazing
9:42 c2[1] will print 'e'
Thank you!
When printing c2[1], why is that printing out a letter? If c2 is a pointer to the address of the first character of c1, wouldn't: print c2[1]; print out 201 (the address of the second element of the array)? I could see how running: print *c2[1]; would give you 'e', but that's not what's written.
Hi! After declaring C1 as character array and C2 as pointer to character, is C2 = C1 somehow the same as C2 = &C1[0] and the compiler does know that now C2 is usable as an array? (since we havent declared it with [] anywhere). Thanks!
bro i love u for this thanks
Muy buenas explicaciones, sin duda!
amazing videos sir.
Thanks 😘
at 4:21 it actually works on my code::block so what is the problem?
You really helped me out a lot!! Thank You!
The code at the end will work...coz when u wrote their *c...u derefrenced a value..named c...in the loop...u printed that derefrenced value....as for the fact that arrays are stored as a reference value in another function and only stores the address of the first variable....now after printing the first variable by drefrencing the address of it ...u incremented it...and in loop u passed the condition not equals to null...so it will terminate their and will give us hello as output...
at 3:52 does it apply for c++???
Good to see such videos....
@3:31 it is a backward slash "\" not forward slash :)
how did" print c2[1] "give the output "l"? isnt it "e" ?
he mentioned in the correction.
Yup
If you increment the c pointer as in the last example. Do you need rewind it back to the original position, else it forever points at the newly incremented address?
Does the same thing which u did in this video also applicable in c++?
at 9:26 , the output should be 'e'
Life saver
Great Stuff!!