Pointers can be hard to learn and understand why or when to use them. This is a nice short video that clearly articulates all the basics in a very clear and easy to follow along example. Thanks for all the videos! Very well done.
I get it now. Pass by reference (pointer) is needed because the variables in functions are exclusive. It's like there's a wall between the function area and the main area. The only way we can connect those variables (in main and in function) is to manipulate the memory via addresses. Kinda like digging a hole underground.😄
See thats the way i want to look at it but, functions can still access variables from the main file without pointers tho?? Like for example: int calculate(int x){ int y = x * 2; return y; Int main(){ int x = 5; int z = calculate(x); printf(“%d”, z); } Like the calculate function still has access to the variable x even without using pointers. So I dont get it.
The key concept here is @@vvvawdad1767, having access to the value or the address of a variable. In your example, the calculate function can only access the value stored in a, as in only *use* a. On the other hand, using pointers, the function could actually *modify* the variable. The use depends on which one you need. Calculate works since it returns a new value that is not related to the argument, however, if you wanted to calculate to change the value of its argument, you would need to use pointers (you also would not need to return anything).
Really well explained, especially giving a reason WHY passing by reference is beneficial. But I am getting confused about *a vs *a. Some times you referred to "*a" like: "a is a pointer" and other times you were saying: "Here we dereference the pointer a". Sounds like "*a" can be two totally different things?
*a = *b *a is an address in memory. *b is the value stored at the address that b points to. So the same notation can mean two different things? Basically I'm totally confused.
@@plica06it’s confusing. Basically the short hand is when you initialize or declare a pointer * is used, eg: Int *a, char *b, float *c Afterwards when you use these pointers * is used to dereference and refer to that actual value at the pointer address. Otherwise, without it you are assigning just the memory address of variables. So in this case: *a=*b the value at a is assigned the value at b a = &b pointer a now points to address of variable b
& = address of... * = access to the value of that address. Using C's good practices, pointers are declared with *pYourVariable and the p stands for 'pointer'. Here is a hard-coded swap that can help you understand: ...hard-coded swap Function... (1) int swap(int *pX, *pY) (2) *pX = 10; (3) *pY = 5 ...main... (1) int x = 5, y = 10; (2) printf(..x..y) //-> x = 5, y = 10 (3) swap(&x, &y); let' say &x = 0x005 and &y = 0x010 are the addresses. (4) printf(..x..y) //-> x = 10, y = 5 Starting from main: (1) declare 'x' equals to 5 and 'y' equals to 10 (2) print the values of 'x' and y, that is 'x' = 5 and 'y' = 10 as in (1) (3) you pass as parameters the addresses of what is stored in 'x' and 'y' (with the &s). (3.1) swap receives the value of those addresses and then it creates the pointers. the value stored in pX is 0x005 and the value stored in pY is 0x010. with the dereference operator, you are accessing to the value of those addresses so *pX is 5 and *pY is 10. (3.2) and (3.3) a hard coded swap, the value of what is stored in the memory address contained in pX now is 10, same for pY. (4) print again the values of 'x' and y, but now 'x' = 10 and 'y' = 5
Then, the swap function showed here should look like this: void swap (int *pX, int *pY) //receive the addresses and get the values inside { int temp; temp = *pX; *pX = *pY; *pY = temp; } int main() { int x = 5, y = 10; printf("'x' is %d and 'y' is %d.", x, y); //-> 'x' is 5 and 'y' is 10 swap(&x, &y); //pass as parameters the addresses where those values are stored printf("'x' is %d and 'y' is %d.", x, y); //-> 'x' is 10 and 'y' is 5 return 0; }
We can do that in C++ where reference variables are supported. But in C we can only achieve 'pass by reference' by using pointer parameters. When we have a parameter like int *a, the argument is going to be a pointer (memory address). :-)
because in C apart from global variables all variables only exist inside the scope of a function. in the example of the video x and y only exist inside the main function and when you pass their values to the swap function it will actually create a copy of them that only exist inside swap, and when the swap function ends the copy will be deallocated and will return to the main function where the actual x and y remained unchanged. so when you actually pass the addresses of x and y as an argument of swap it'll change their values for the whole program, because the actual value stored in the memory address of x and y will be changed other than a copy that only exists in the scope of swap
Great question Bruno! :-) The term pass by reference is more accurate in languages like C++ that have reference variables (and reference parameters): ruclips.net/video/cxysUPZH65Y/видео.html. Reference variables are a different type of variable than a pointer, though the concepts are a bit similar too... references are a reference to the variable, the essentially "are" the variable they are referencing and give us access to that variable. Whereas pointers store a memory address, the memory address of the variable they are "pointing to", and we can access the variable they are "pointing to" by de-referencing the pointer. So in C, because we are using pointers, it's more accurate to say "pass by pointer". As a practical matter though if you say "pass by reference" when referring to "pass by pointer" people will know what you mean and the terms get used in a more casual manner like that in practice.
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Pointers can be hard to learn and understand why or when to use them. This is a nice short video that clearly articulates all the basics in a very clear and easy to follow along example. Thanks for all the videos! Very well done.
i absolutely love this C tutorial series.
your channel is amazing! i need to pick up the pace, i still got 154 videos left to grind..
I’m really glad that you’re enjoying the content. :-)
This concept is hard to get for beginners, thanks a lot!
You're very welcome Philipp! :-)
Watched so many videos on this. First time I actually understood what it is. Good video.
I get it now. Pass by reference (pointer) is needed because the variables in functions are exclusive. It's like there's a wall between the function area and the main area. The only way we can connect those variables (in main and in function) is to manipulate the memory via addresses. Kinda like digging a hole underground.😄
See thats the way i want to look at it but, functions can still access variables from the main file without pointers tho??
Like for example:
int calculate(int x){
int y = x * 2;
return y;
Int main(){
int x = 5;
int z = calculate(x);
printf(“%d”, z);
}
Like the calculate function still has access to the variable x even without using pointers. So I dont get it.
The key concept here is @@vvvawdad1767, having access to the value or the address of a variable. In your example, the calculate function can only access the value stored in a, as in only *use* a. On the other hand, using pointers, the function could actually *modify* the variable. The use depends on which one you need. Calculate works since it returns a new value that is not related to the argument, however, if you wanted to calculate to change the value of its argument, you would need to use pointers (you also would not need to return anything).
Passed by pointer. Got it!!!
Great video, thanks!
You're welcome! :-D
pass by pointer 💪thnx a lot, this channelis realy one of the best c channel ever :)
Haha thanks I'm glad you think so Muhi! :-)
This is gold!!! Thank you for the clear explanation
You're very welcome! :-)
The Best channel so far
Thank you very much Louis! :-)
Clear as a whistle
Really well explained, especially giving a reason WHY passing by reference is beneficial. But I am getting confused about *a vs *a. Some times you referred to "*a" like: "a is a pointer" and other times you were saying: "Here we dereference the pointer a". Sounds like "*a" can be two totally different things?
*a = *b
*a is an address in memory.
*b is the value stored at the address that b points to.
So the same notation can mean two different things?
Basically I'm totally confused.
@@plica06it’s confusing. Basically the short hand is when you initialize or declare a pointer * is used, eg:
Int *a, char *b, float *c
Afterwards when you use these pointers * is used to dereference and refer to that actual value at the pointer address. Otherwise, without it you are assigning just the memory address of variables. So in this case:
*a=*b the value at a is assigned the value at b
a = &b pointer a now points to address of variable b
& = address of...
* = access to the value of that address.
Using C's good practices, pointers are declared with *pYourVariable and the p stands for 'pointer'. Here is a hard-coded swap that can help you understand:
...hard-coded swap Function...
(1) int swap(int *pX, *pY)
(2) *pX = 10;
(3) *pY = 5
...main...
(1) int x = 5, y = 10;
(2) printf(..x..y) //-> x = 5, y = 10
(3) swap(&x, &y); let' say &x = 0x005 and &y = 0x010 are the addresses.
(4) printf(..x..y) //-> x = 10, y = 5
Starting from main:
(1) declare 'x' equals to 5 and 'y' equals to 10
(2) print the values of 'x' and y, that is 'x' = 5 and 'y' = 10 as in (1)
(3) you pass as parameters the addresses of what is stored in 'x' and 'y' (with the &s).
(3.1) swap receives the value of those addresses and then it creates the pointers. the value stored in pX is 0x005 and the value stored in pY is 0x010. with the dereference operator, you are accessing to the value of those addresses so *pX is 5 and *pY is 10.
(3.2) and (3.3) a hard coded swap, the value of what is stored in the memory address contained in pX now is 10, same for pY.
(4) print again the values of 'x' and y, but now 'x' = 10 and 'y' = 5
Then, the swap function showed here should look like this:
void swap (int *pX, int *pY) //receive the addresses and get the values inside
{
int temp;
temp = *pX;
*pX = *pY;
*pY = temp;
}
int main()
{
int x = 5, y = 10;
printf("'x' is %d and 'y' is %d.", x, y); //-> 'x' is 5 and 'y' is 10
swap(&x, &y); //pass as parameters the addresses where those values are stored
printf("'x' is %d and 'y' is %d.", x, y); //-> 'x' is 10 and 'y' is 5
return 0;
}
Got it ! thank you a lot !
Excellent, and you're welcome! :-)
Can you have (int &a, int &b) as function perameters? Im confused here? As we want an address to deference...?
We can do that in C++ where reference variables are supported. But in C we can only achieve 'pass by reference' by using pointer parameters. When we have a parameter like int *a, the argument is going to be a pointer (memory address). :-)
thx man I immediately understood everything after watching this video😁
still didnt understand why call by value didn't work...
😆
but I understood the call by reference mechanism,
thanks!!
because in C apart from global variables all variables only exist inside the scope of a function.
in the example of the video x and y only exist inside the main function and when you pass their values to the swap function it will actually create a copy of them that only exist inside swap, and when the swap function ends the copy will be deallocated and will return to the main function where the actual x and y remained unchanged.
so when you actually pass the addresses of x and y as an argument of swap it'll change their values for the whole program, because the actual value stored in the memory address of x and y will be changed other than a copy that only exists in the scope of swap
really good, thanks you!:)
You're welcome! :-)
Really thanks ❤
You’re welcome! ❤️
thank you do much such a great video.
You're welcome Yoeum, I'm glad you enjoyed it! :-)
cool man!
thx
You’re welcome! :-)
if we think that & is inverse of *, & times * will produce 1. Therefore *&x = x. (i dont know is that interpretation whether correct or not)
No
what IDE you're using sir?
and thanks for the video.
You're welcome! :-) In this video I am using Visual Studio Code as a text editor, and I am using the gcc compiler on the MacOS Terminal.
why is it called passed by pointer and not passed by reference?
Great question Bruno! :-) The term pass by reference is more accurate in languages like C++ that have reference variables (and reference parameters): ruclips.net/video/cxysUPZH65Y/видео.html. Reference variables are a different type of variable than a pointer, though the concepts are a bit similar too... references are a reference to the variable, the essentially "are" the variable they are referencing and give us access to that variable. Whereas pointers store a memory address, the memory address of the variable they are "pointing to", and we can access the variable they are "pointing to" by de-referencing the pointer. So in C, because we are using pointers, it's more accurate to say "pass by pointer". As a practical matter though if you say "pass by reference" when referring to "pass by pointer" people will know what you mean and the terms get used in a more casual manner like that in practice.
Godsent
Ahhhhh i still dont get it but it kinda helped