Character arrays and pointers - part 1
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- čas přidán 8. 07. 2024
- See complete series on pointers in C/C++ here:
• Pointers in C/C++
See part 2 of this lesson here:
• Character arrays and p...
In this lesson, we have described how we can work with character arrays using pointers. Character arrays are used to store strings in C and we work with them for all kind of string manipulation. Working with string in C is tricky and requires good understanding of pointers.
Feel free to drop your comments. feedback and suggestions.
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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.
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.
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
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!
@mycodeschool ur tutorials are simply the best. thx a lot
You have made a superb video - many thanks.
Thanks for your tutorial! You're the best!
These vidz are fantastic thanks for your time creating and uploading them...
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
Awesome! God bless you. Good example when you first didn't insert and then inserted null character at the end of string.
brilliant !
Very easy and efficient explanations.
Thank you so much sir.
you are such a great teacher, thanks a lot! your video helped me hugely!
Thank you very much admin you taught really classical
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
Great video's series on pointers. You make it very clear.
thank you!
Seriously helpful for me to understand this "array" thingy, thanks dude for the video
Really Awesome, great and very clear understanding of the concepts! Thank you very much!
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 :)
Excellent videos in general, the best I've seen so far.
thanks it helped me in passing character array as argument
Amazing video!
Amazing Video which is mastered by your hard work
Kudos!!!!
thank you for the time you take to make these amazing videos!
very nice tutorial.
thanks a lot.
Thanks for the `while` idea, solved a problem I was suffering from.
Muy buenas explicaciones, sin duda!
Many many many thanks .... You was very helpfull for me ... Your way in explaining things is more than excellant .... Thank you
thanks for your explaination!
Excellent video with the best explanation I have seen yet, thank you!! Plus leaving us homework :)
these videos are great sir.. i am recommending this channel to my friends and teacher..
Thank you for clearing my doubts
your videos are amazing
these videos are gold
amazing explaining, thank you :)
awesome work sir, really appreciating, thank you 👍👍
are Deepak sir
Good quality and depth.
Thank you!
Many thanks
And. We loved this explanation!👌
thankyou so much, your videos are so convincing :-)
thank u very much you are simply amazing
perfect explanation
Your videos saved my exams at university
Thank you from Greece!
Well it saved you not your exams lol
This is gold.
Great Stuff!!
thank you for the great lesson . i searched these information from you tubel a lot.
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.
Amazing!
I believe that in a list the null terminated character is implicitly declared ?
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 :)
amazing videos sir.
Thank you very much! I finally find my bug after watching it.
thanks a lot sir...
Life saver
Good to see such videos....
Very beautiful handwriting ... Nice explanation too
Thank you
Awesome video for beginner like me ...
this may help me for passing this part
great video
OO man , this man is doing magic even after 10 years
thank you sir
Thanks 😘
You really helped me out a lot!! 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
bro i love u for this thanks
nice video!
Hi,
Could you suggest a list of exercices for pointers?
thank you
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.
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)
u really worked hard. thank u for uploading. it really helped me a lot.
i can use pointer like a expert.
Increementing of C works because. in print function it is reference value which can be incremented
These videoes really help me.
why my compiler shows the whole char array instead of prints out the first character of the array when i use cout to print C2?
Thanks very useful for me
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?
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 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?
How does printing the string of character array with %s format work since arrays point to the first element of the array??How does the array name for the string go to all the different addresses of the elements??
please make more videos on programming
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
very good video .You explain it beautifully . Other series kb ay gi ???
is it possible to declare a character array that will accept input without placing size to make it a variable size array or you have to explicitly set the size. by that i mean there is no initialization just declaration of it. so it can be used later on as the program executes to take user input
very well handed
do you have any youtube resources for c++ beginners covering the technical under the hood stuff, that dont just superficially walking you through writing a program?
what compiler are you used?
u r amazing sir!!
no he is not amazing......he made a lot of mistakes
@@sanambaloch2975 like to share those mistakes?
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.
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.
Which compiler are u using?
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
Sweet
nice! :)
Lifesaver
sir which C compiler are you using
sir we can only icrement pointer variable only when it is assigned to other but how can we increment c.?
Dear author and all,
Why the output of program without character '\0' is ok, didn't see same your result?
I am using ubuntu 15.1 and g++ compiler.
thank!
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.