linker links ur obj files into executable file

I'm really sorry, but this is what Google says too, and I'm not sure I follow. What I don't understand is simple : let's say that you have a source file(A.cpp), you include certain files of interest in it and then simply compile that file(A.cpp) to get your executable. why do we need anything else? I've thought about it and I'm lost.

I'm really sorry, but this is what Google says too, and I'm not sure I follow. What I don't understand is simple : let's say that you have a source file(A.cpp), you include certain files of interest in it and then simply compile that file(A.cpp) to get your executable. why do we need anything else? I've thought about it and I'm lost.

This way, we would need to compile only a single file for all the projects.

i want an output like e em eme emen emenp emenpl emenpla emenplac

And hence, why do we need makefiles too?

ur cpp files translate into machine language (machine units), files only with bites, then into obj files, then linto linker

Let's say we have a file(A.cpp) which has only the content #include<iostream> int main(){cout<<"hi";} Could you please walk me through it's compilation process ? Thanks.

iostream is a file in the computer, whose path the compiler already knows. The compiler transforms the .cpp (source) in a .o (object), then the linker transforms the .o file in the final binary

iostream is a file in the computer, whose path the compiler already knows. The compiler transforms the .cpp (source) in a .o (object), then the linker transforms the .o file in the final binary

iostream is a file in the computer, whose path the compiler already knows. The compiler transforms the .cpp (source) in a .o (object), then the linker transforms the .o file in the final binary

iostream is a file in the computer, whose path the compiler already knows. The compiler transforms the .cpp (source) in a .o (object), then the linker transforms the .o file in the final binary

Is iostream a .cpp file?

But I thought the linker linked files.

linker links ur object files only into 1 exe file

linker links ur object files only into 1 exe file

Can we say it as a simple conversion only? That it converts object file to an executable?

But I thought the linker linked files.

When you include a header file, you just get the API, but the actual implementation of that API may be in another already compiled file, which is linked later

So including is not a simple recursive copying of the file contents to the source file, right ?

So including is not a simple recursive copying of the file contents to the source file, right ?

Let's think differently. If you declare a function with no body, the compiler is fine with it, as long as you provide an object file or a library with the implementation to the linker

Is library an object file?

Not exactly, but close enough

Libraries are essentially object files that are linked in a clearer way

Let's think differently. If you declare a function with no body, the compiler is fine with it, as long as you provide an object file or a library with the implementation to the linker

I'm really sorry, but do you mean that including a file doesn't simply mean copying its content to the source file.

So you compile the source only once and not everytime you compile the main file

But theoretically it is possible to write a single file cpp file for all the projects, but it would be highly unadvised because of modularity issues, am I correct?

Yes.

#include<bits/stdc++.h> using namespace std; int main(){ vector<pair<string,pair<int,int>>> v; v.push_back(make_pair("hello",make_pair(3,5))); v.push_back(make_pair("world",make_pair(5,6))); for(auto element:v){ cout<<element.first<<endl; cout<<element.second.first<<" "; cout<<element.second.second<<endl; } sort(v.begin(),v.end(),compare) return 0; }

#include<bits/stdc++.h> using namespace std; int main(){ vector<pair<string,pair<int,int>>> v; v.push_back(make_pair("hello",make_pair(3,5))); v.push_back(make_pair("world",make_pair(5,6))); for(auto element:v){ cout<<element.first<<endl; cout<<element.second.first<<" "; cout<<element.second.second<<endl; } sort(v.begin(),v.end(),compare) return 0; }

Why not using tuple<string, int, int> instead? That's not a very friendly snippet of code

Actually it is given and mandatory ...can you tell me how to do this?

Including a file literally means copying the content of one file in another

I'm sorry, but another question: when including means copying the contents of a file into main file, then, let's say a file(A.cpp) includes iostream, and iostream includes ios, does it mean that the code of ios too is copied to the main file?

Nobody wants a file with 50.000 lines of code

Yes

So eventually what happens is the main file becomes very lengthy.

And when the main file becomes too lengthy, doesn't it mean that the code of iostream ends up being compiled again? Because now, it is included in the main file itself?

Also iostream is template based, so you have no other options

Also iostream is template based, so you have no other options

You got it well enough 👍

#include <iostream> #include <cmath> using namespace std; int check(int n); int check(int n) { if (n=0) return 0; if (n=1) return 1; if (n=2) return 2; return floor(check(n/2))+floor(check(n/3))+floor(check(n/4)); } int main() { int n; int gift; cin>>n; while(n!='\0') { gift=check(n); cout<<gift<<endl; cin>>n; } return 0; }

I am trying to learn Dynamic Programming and cant quite figure out where I'm going wrong...

equality sign

equality sign

i changed that.. still got some problem.. is that long return statement with floor permitted?

#include <iostream> #include <cmath> using namespace std; int check(int n); int check(int n) { if (n=0) return 0; if (n=1) return 1; if (n=2) return 2; return floor(check(n/2))+floor(check(n/3))+floor(check(n/4)); } int main() { int n; int gift; cin>>n; while(n!='\0') { gift=check(n); cout<<gift<<endl; cin>>n; } return 0; }

and there can be any number of inputs

#include <iostream> #include <cmath> using namespace std; int check(int n); int check(int n) { if (n=0) return 0; if (n=1) return 1; if (n=2) return 2; return floor(check(n/2))+floor(check(n/3))+floor(check(n/4)); } int main() { int n; int gift; cin>>n; while(n!='\0') { gift=check(n); cout<<gift<<endl; cin>>n; } return 0; }