What the fu** are you talking about? Do you even know what a freestanding implementation is? It doesn't mean your own implementation. The C and C++ standards define two implementation types - hosted and freestanding. https://en.cppreference.com/w/cpp/freestanding Freestanding implementation is what Ammar Faizi was talking about in his original post that you and someone else were arguing about. And you said size_t and uintptr_t won't be available on those implementations. And I just proved you wrong. Don't argue without understanding what someone is trying to say.

Then compile -nostdlib code with stdint.h header. Prove it in practice

It means I cannot use coroutines in freestanding because libstdc++ doesn't support the clang intrinsics yet

Then compile -nostdlib code with stdint.h header. Prove it in practice

I... I do that everyday

Coroutines are not meant to be user facing yet. The C++20 standard just defined the backbone structure. With C++23, full fledged support will be available and freestanding implementations hopefully will offer full support by then

Ok. I thinked that stdint.h has libc dependent macroses. I am wrong ok

Ok. I thinked that stdint.h has libc dependent macroses. I am wrong ok

It uses stddef tho iirc, which is also freestanding-OK

It uses stddef tho iirc, which is also freestanding-OK

Is uintptr_t and size_t has different size in Real Mode?

Is uintptr_t and size_t has different size in Real Mode?

Is uintptr_t and size_t has different size in Real Mode?

Casting from pointer to integer and vice versa should use uintptr_t or intptr_t. Not size_t. Because the pointer size can be bigger than the size_t. For example: C compilers on x86 real mode can have 32 bit FAR or HUGE pointers but size_t is still 16 bits.

I asked about that case

Emm... what did you ask?

Is it really that size_t and uintptr_t can be not same

Is it really that size_t and uintptr_t can be not same

On systems with far and near pointers, pointers are generally 2*size_t in size

int getX() const {return x;} what is the purpose of const in this line?

It means this method won't change the object pointed to by this pointer. The this pointer will be a pointer to a const in this method. So you can call this method on a const object. If you remove the const, you can't call this method using a const object

In this line of code printf("%2x", opc[i] & 0xFF);, what does the & operator do?

In this line of code printf("%2x", opc[i] & 0xFF);, what does the & operator do?

Binary conversion

Can you look that Post? https://stackoverflow.com/q/74089335/14686302

binary conversion? O.o

Bitwise operators perform functions bit-by-bit on either one or two full binary numbers.

I believe "Introduction to Algorithms, 3rd Edition" by Thomas H. Cormen has relevant chapters.

Hi,I have a doubt about arrays: does it exist a way to pass a 2D array to a function with at least one variable dimension? For instance,I want the users to type the dimensions and then I would pass the array to a function

This is wrong. ptr[x] is *(ptr+x)

Yes, https://t.me/programminginc/441139

And please if you have any sources of exercises … about time complexity big o notation , For more practice ..

why i input p=helloworld and sizeof(p)=8 ?

Yes

Pointer is always 8 byte long... if you want to get the length of the string use strlen(str);

char *name[]={"Follow me"}

#include <stdio.h> #include <string.h> int main() { char a[15] = "Hello world"; int l = strlen(a); printf("Length:%d",l); }

What if i have to use char *a 🥲

#include <stdio.h> #include <string.h> int main() { char a[] = "Hello world"; char *b = a; int l = strlen(b); printf("Length:%d",l); } Where a is the array of char and b is pointer to that array.

ll solve(){ ll n; cin>>n; ll loo=n; ll a[n]; for(int i=0;i<n;i++){ a[i]=i+1; } cout<<n<<"\n"; while(loo>0){ cout<<n<<"\n"; for(int i=0;i<n-1;i++){ cout<<i+1<<a[i]<<a[i+1]<<"\n"; } cout<<n<<a[0]<<a[n-1]<<"\n"; loo--; } }

What type of data allows you to make a = 1.5 a % 1 = 0.5? in C

none... you would need to use functions for that

Thanks

what's ll?

https://hastebin.com/okotoyujez.cpp

ll solve(){ ll n; cin>>n; ll loo=n; ll a[n]; for(int i=0;i<n;i++){ a[i]=i+1; } cout<<n<<"\n"; while(loo>0){ cout<<n<<"\n"; for(int i=0;i<n-1;i++){ cout<<i+1<<a[i]<<a[i+1]<<"\n"; } cout<<n<<a[0]<<a[n-1]<<"\n"; loo--; } }

How int a => double b?

#include <stdio.h> #include <string.h> int main() { char a[] = "Hello world"; char *b = a; int l = strlen(b); printf("Length:%d",l); } Where a is the array of char and b is pointer to that array.

#include <stdio.h> int main(void) { unsigned int l = 0; char a[] = "Hello world"; char *b = a; while(*b != '\0') l++, b++; printf("Length: %d", l); return 0; } Using pointer arithmetic we can do the same thing without depending on the string header file.

You can find many solutions to the same problem in C.

in fact infinitely many solutions 😊 but some are more reasonable than others