error: no matching function for call to 'std::multiset<int>::erase(std::multiset<int>::reverse_iterator&)' I'm getting this

You're calling erase with reverse iterator

You're calling erase with reverse iterator

oh !! so erase only works on normal iterator and not reverse iterators, right?

oh !! so erase only works on normal iterator and not reverse iterators, right?

Yes

There are 2 methods to remove the last element minheap.erase(std::prev(minheap.end())); and minheap.erase(--minheap.end()); Is there any preferred one?

Hello guys, I just wanted to know where I can learn C online. All suggestions are welcome, thanks! @Rtr_238

If you are a beginner, then start with hackerrank and then move forward to other coding platforms like Codechef, Codeforces. And, if you want to learn advanced concepts like dp, graph then u should refer to gfg. You can also use gfg as a beginner.

There are 2 methods to remove the last element minheap.erase(std::prev(minheap.end())); and minheap.erase(--minheap.end()); Is there any preferred one?

I'd use first

If you are a beginner, then start with hackerrank and then move forward to other coding platforms like Codechef, Codeforces. And, if you want to learn advanced concepts like dp, graph then u should refer to gfg. You can also use gfg as a beginner.

This is a bad way

No need to mutate stuff

I'd use first

+1

I'd use first

and the possible reason is ......

and the possible reason is ......

Looks better

and the possible reason is ......

https://t.me/programminginc/307745

https://t.me/programminginc/307745

didn't get the mutate word that he described 😅

didn't get the mutate word that he described 😅

--end() — mutating an iterator

--end() — mutating an iterator

🆗 now, makes sense

didn't get the mutate word that he described 😅

To put it short mutating is calling non-const method or assigning

Ops, it does

class node { node(int value) }; // What is the meaning of node(int value) ?

But this mutation is hidden for best

class node { node(int value) }; // What is the meaning of node(int value) ?

Its a constructor

Caller doesnt care how prev gets previous iterator

#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> int main() { int n,i,j,k; scanf("%d", &n); for(i=1;i<=(n*2)-1;i++) { for(j=1;j<=(n*2)-1;j++) { if(i==1||i==(n*2)-1||j==1||j==(n*2)-1) { printf("%d ",n); } else { printf("%d ",n-1); //++; } } printf("\n"); } return 0; }

help me with this

lmao

read the rules

What's the difference between std::int64_t, long and long long?

They all are 64 bits in size and represent same thing.

Long size depends on architecture

They all are 64 bits in size and represent same thing.

Sizes of long and long long are depends on the architecture https://en.cppreference.com/w/cpp/language/types std::int64_t is an alias for a type that is 64 bit on your current architecture

Sizes of long and long long are depends on the architecture https://en.cppreference.com/w/cpp/language/types std::int64_t is an alias for a type that is 64 bit on your current architecture

ok so if I have 32 bit system will their size become 32 bits?

Sizes of long and long long are depends on the architecture https://en.cppreference.com/w/cpp/language/types std::int64_t is an alias for a type that is 64 bit on your current architecture

sorry I can't access browser atm, Ram almost full

ok so if I have 32 bit system will their size become 32 bits?

As far as I can see, long will be 32 and long long will be still 64. As it said in the docs long long need to be at least 64-bit. And it seems that there's no architecture where long long is bigger, but I'm not sure about this.

As far as I can see, long will be 32 and long long will be still 64. As it said in the docs long long need to be at least 64-bit. And it seems that there's no architecture where long long is bigger, but I'm not sure about this.

hmm... I see

as no 128 bit architecture cpu exists so long long can't be changed

Sizes of long and long long are depends on the architecture https://en.cppreference.com/w/cpp/language/types std::int64_t is an alias for a type that is 64 bit on your current architecture

so in my case whose alias is std::int64_t exactly is?

I have a 64 bit system

so in my case whose alias is std::int64_t exactly is?

although that doesn't really matter, but question is what should I use and in what circumstances?

I have a 64 bit system

I'm not sure if it is specified exactly in case these two are types of the same size https://en.cppreference.com/w/cpp/types/integer signed integer type with width of exactly 8, 16, 32 and 64 bits respectively with no padding bits and using 2's complement for negative values (provided only if the implementation directly supports the type) (typedef)

So it can be either long or long long in your case, I guess you can find it in your implementation

So it can be either long or long long in your case, I guess you can find it in your implementation

hmm...

I'll check it later

although that doesn't really matter, but question is what should I use and in what circumstances?

what about this?

like should I use std::int64_t or long long?

or if it depends then how exactly are they supposed to be used

like should I use std::int64_t or long long?

In my opinion, if the number is going to be big (and that's probably why you choose long instead of int), then it's better to use std::int64_t so you know exactly how much data you can fit in it not depending on the platform.

In my opinion, if the number is going to be big (and that's probably why you choose long instead of int), then it's better to use std::int64_t so you know exactly how much data you can fit in it not depending on the platform.

Hmm... thank you very much for clearing my doubt there Kudos

but do note that [u]int<SZ>_t types are optional

but do note that [u]int<SZ>_t types are optional

you mean I shouldn't do that?

I think based on what @gameraccoon said, I think its okay to say that std::int8_t is char

but do note that [u]int<SZ>_t types are optional

oh I saw the edit, thanks

just wanted to point out, there are (have been) systems that used 5-, 6-, 7-, and 9-bit bytes - on those platforms int8_t would not be available.

just wanted to point out, there are (have been) systems that used 5-, 6-, 7-, and 9-bit bytes - on those platforms int8_t would not be available.

oh, that's new to me, then those systems must be really old I guess

what

wtf?

Click here /rules

LMAO

help me with this

why

help me with this

no please?

/ban Siku T

well I'm pretty sure that I'll be using 64 bit type, but int_fast64_t seems to lose precision isn't it?

and if it goes low?

then what makes it different from int64_t?

actually I'm using constructors of different types and let compiler decide the type. see image.

Hmm... I understood, but I don't see the point targeting those machines

I'm solely making it for myself and that's a 64 bit system

yes, thanks

alright then time to continue on

actually I'm using constructors of different types and let compiler decide the type. see image.

Then why do not use templates? Let compiler do it for you.

Then why do not use templates? Let compiler do it for you.

I can but each constructor does different thing depending on type

with templates I can't achieve that

I can but each constructor does different thing depending on type

Same algorithm but different thing or different algorithms depending on type?

Same algorithm but different thing or different algorithms depending on type?

different algorithm depending on type

I can but each constructor does different thing depending on type

Are u from china?

Are u from china?

What makes you think so?

What makes you think so?

Your name

Maths Lover

You could have translated that with gtranslate or something, but please avoid any furthur conversation of such manner as it's off topic and should be continued in OT group. Thank you for understanding

Fuck china

Fuck china

lol