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.
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
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 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
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.
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.
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 :)
@@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.
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.
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.
*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
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
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.
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
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++
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.
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.
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.
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)
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!
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
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.
I see everybody is benefiting from your teaching, and I will thank you for doing so. Great simple concise and complete explanation. Thank you
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.
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
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.
this channel is just amazing, up there with the best cs channels in the world
This video didn't become the top viewed in this subject for nothing . Thank you sir
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
Your videos saved my exams at university
Thank you from Greece!
Well it saved you not your exams lol
OO man , this man is doing magic even after 10 years
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!
You have made a superb video - many thanks.
@mycodeschool ur tutorials are simply the best. thx a lot
Seriously helpful for me to understand this "array" thingy, thanks dude for the video
awesome work sir, really appreciating, thank you 👍👍
are Deepak sir
you are such a great teacher, thanks a lot! your video helped me hugely!
Awesome! God bless you. Good example when you first didn't insert and then inserted null character at the end of string.
Many many many thanks .... You was very helpfull for me ... Your way in explaining things is more than excellant .... Thank you
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 :)
Thank you very much admin you taught really classical
These vidz are fantastic thanks for your time creating and uploading them...
thank you for the great lesson . i searched these information from you tubel a lot.
thank you for the time you take to make these amazing videos!
Great video's series on pointers. You make it very clear.
thank you!
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.
Thanks for the `while` idea, solved a problem I was suffering from.
Thanks for your tutorial! You're the best!
Thank you for clearing my doubts
u really worked hard. thank u for uploading. it really helped me a lot.
i can use pointer like a expert.
these videos are great sir.. i am recommending this channel to my friends and teacher..
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 :)
Excellent video with the best explanation I have seen yet, thank you!! Plus leaving us homework :)
Very beautiful handwriting ... Nice explanation too
Excellent videos in general, the best I've seen so far.
thanks for your explaination!
these videos are gold
brilliant !
Very easy and efficient explanations.
Thank you so much sir.
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.
perfect explanation
Wonderful. You’re so awesome.
thanks it helped me in passing character array as argument
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!!!!
Really Awesome, great and very clear understanding of the concepts! Thank you very much!
this may help me for passing this part
Many thanks
at 3:33 its backward slash
This is gold.
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
Good quality and depth.
Awesome video for beginner like me ...
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.
*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
very nice tutorial.
thanks a lot.
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
Amazing video!
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.
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 :-)
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
And. We loved this explanation!👌
bro i love u for this thanks
Thanks 😘
Thank you
your videos are amazing
You really helped me out a lot!! Thank You!
Thank you very much! I finally find my bug after watching it.
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
very good video .You explain it beautifully . Other series kb ay gi ???
great video
Thank you!
thank you sir
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?
Increementing of C works because. in print function it is reference value which can be incremented
Good to see such videos....
amazing videos sir.
C++ will work, because you are incrementing the start address that was given as a parameter!
thanks a lot sir...
Because C is a pointer to a character. And C++ means to increment the value of C by 1 byte (the address)
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.
thank u very much you are simply amazing
amazing explaining, thank you :)
I believe that in a list the null terminated character is implicitly declared ?
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.
Great Stuff!!
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!
Thanks very useful for me
Muy buenas explicaciones, sin duda!
you are god to me
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)
Life saver
9:42 c2[1] will print 'e'
at 9:26 , the output should be 'e'
please make more videos on programming
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!
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
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