what is neovim

middle school (in fact

You should pass the array size too, like: bool f(int* ptr, size_t sz) { /* ... */ } The name of an array will decay to a pointer too much easily but you need the size in C++ and C (I assume). One of the advantages of using C++ over C is its brilliant STL containers, like std::array or std::vector in this case, which include their size with themselves.

We can do it without passing size if I do *(ptr+i) !='\0' Even I am doing memory allocation dynamically it's telling The correct length i have taken 0 and length=0 and using while(1) till it is not null character it will do i+1 and length+1

C-style array is not null terminated.

If the size is not mentioned explicitly, then the null terminator is included at the end automatically.

It is the string literal which is null terminated not an array of any type.

What is your code?

Exception: if you're talking about argv, it is indeed always a null-terminated array. https://stackoverflow.com/a/11020198

And envp as the third possible parameter to main is also null terminated i suppose

my course uses Dev-C++, and I still use my neovim to take the course

Isn't envp not defined by C/C++ standard?

If size of an array is also given and ptr that points to that array given in a function then how we can do

func(void *ptr, size_t nbyte)

But I have done without take size By pointer arithmetic making length+ until *(ptr+i) is not equal to null character

But I have done without take size By pointer arithmetic making length+ until *(ptr+i) is not equal to null character

if (i) in C/C++ is basically if (i != 0) or if (i != '\0')

Of?

Do you mean 'or'? Please use question marks if you are questioning something

if (i) in C/C++ is basically if (i != 0) or if (i != '\0')

IMHO, if(i) is a lazy way to say "you are dumb if you don't understand it". When you make an if statement, the answer can only be zero or non-zero. For example, when using an specific value, you explicitly express the value: if ( my_value == 5 ) Or in the case of chars: if ( the_element_of_that_string == 'a' ) But, the "i am smarter than everybody" kind of people uses hacks (yes, its literally a hack, a code cut), and, when you want to check the value explicitly, for example, with pointers, you could express in that way: if ( my_pointer == NULL ) But no, the usual approach is the implicit if capability (your programming teacher doesn't tell anything about). There is an implicit conversion to a "true/false" state, so, if you use the "hacky" way (remember the dictionary definition of hack: a cut) if ( my_pointer ) The if itself adds for you (even if you don't want to) the internal conversion to a "true/false". Because that means explicitly "non-zero/zero", the value of the variable you can enter there is checked specifically for zeroes -> false case, and whatever other value -> true

if (i) in C/C++ is basically if (i != 0) or if (i != '\0')

So, the answer to your question "Is the != 0 or != '\0'" is both. That's the magic! Because '\0' is de char ASCII representation of a 0x00 value in 8 bits, and the 0 is the representation of (in 32 bit machine) 0x00000000. So, strictly are not exactly the same (because of the types), but they are exactly the same (at value comparison): both compares with the non-zero/zero.

Actually I refer to this message https://t.me/programminginc/501268. > By pointer arithmetic making length+ until *(ptr+i) is not equal to null character But thanks for the additional information anyway :D

Like this? void foo(char* charPtr) { for(*charPtr; *charPtr != '\0'; ++charPtr) // do something with the element } int main() { char a[] {"abcd"}; foo(a); }

Ya but the array i have taken was type int And used while loop

So where's the problem?

Now there is no problem I was facing how to find the length at morning when I was solving questions in leatcode it Haven't give me size of array and was asking to sort that's why i was asking to how to get length of array using pointers . By this learned a. New way to solve it

Didn't you know for (i = 0; i < X; i++) is same with i = 0 while (i < X) i++

I know this we can do this if size is given

The reason why strcpy doesn't need length as strncpy because it can find the end of string limit by finding the null terminator

But the array i have taken was int I will give u screenshot of code at night when I will open my laptop That's why i was confused how to do it

Then why with int ? It just different data type that has different size

If you want to convert into byte just multiple it with sizeof(int)

See the question was this U have given an array test cases were by leatcode terminal only Not mine So the array was given by that testcase and i have to sort it but the problem was that I have only access to pointer that's pointing that's array first element. I was not knowing where to terminate thats it that's why i ask how to know length of array using pointers So my approach was that I do pointer arithmetic and find length till*(ptr+i) is not equal to null character But in int array how can null character be there i was confused here 🧐

int i[1] is basically char i[4]

Hm because size of int is 4bit or 2 depend on machine

What I'm trying to explain to you that in memory it can be everything

Ohk

Have you learn about heap memory allocation?

By malloc calloc

Did you know what types are that malloc() and calloc() return ?

Void pointers

May be because we are allocation memory dynamically and we can change it to any other pointer too Because void is a generic pointer can be cast to any other pointer

Yep, that's also can be applied when searching for terminators in array

It doesn't matter what its types. It just what terminators you define to indicate the end of array.

If possible, point me to a good resource on internet

If possible, point me to a good resource on internet

If possible, point me to a good resource on internet

If possible, point me to a good resource on internet

@asdfqwerty1234567890 these are string manipulation