how can i call a function during runtime? for example i run the code and input show then output an integer

how can i call a function during runtime? for example i run the code and input show then output an integer

Do all these actions in the same loop in order I think

Do all these actions in the same loop in order I think

You're right

Do all these actions in the same loop in order I think

what do you mean

Hello there, can someone explain how line 27 works in this code (https://gist.github.com/smthinthewayy/8b130b64b033e51e6cad8e640ee70ebf).

eg log(123) = 2 * log(10) + 0 * log(2) + log(0.229)

how?))

it just multiplies and adds some numbers?

it just multiplies and adds some numbers?

yep, number of divisions by 10 * log(10) + number of divisions by 2 * log(2) + log(x - 1)

eg log(123) = 2 * log(10) + 0 * log(2) + log(0.229)

123 / 10 == 12.3, first_part = 1 12.3 / 10 == 1.23, first_part = 2 1.23 < 2, second_part = 0 gets 2 * log(10) + 0 * log(2) + log(1.23 - 1 == 0.3)

it works perfectly, but I don't understand how, what is this property?

Please what's are the variables in c programming Can only get 5 from net

#include <iostream> using namespace std; int main(int argc, char** argv) { int x,y,z; cout <<"Input Number 1: "; cin >> x; cout <<"Input Number 2: "; cin >> y; cout <<"Input Number 3: "; cin >> z; if (x=y=z){ cout<<"all 3 equal"; } else cout<<"not all equal"; return 0; } What is wrong with this code? I'm getting same output

#include <iostream> using namespace std; int main(int argc, char** argv) { int x,y,z; cout <<"Input Number 1: "; cin >> x; cout <<"Input Number 2: "; cin >> y; cout <<"Input Number 3: "; cin >> z; if (x=y=z){ cout<<"all 3 equal"; } else cout<<"not all equal"; return 0; } What is wrong with this code? I'm getting same output

= is different with ==

#include <iostream> using namespace std; int main(int argc, char** argv) { int x,y,z; cout <<"Input Number 1: "; cin >> x; cout <<"Input Number 2: "; cin >> y; cout <<"Input Number 3: "; cin >> z; if (x=y=z){ cout<<"all 3 equal"; } else cout<<"not all equal"; return 0; } What is wrong with this code? I'm getting same output

You should use == to declare condition, single = is for declaring variables😁👍

You should use == to declare condition, single = is for declaring variables😁👍

> = is for declaring variables = is an assign operator

You can define variables later, not at declaration time

Use &&

x == y later will become 0 or 1 (result of comparing x and y), and then become 0 == z or 1 == z

x == y later will become 0 or 1 (result of comparing x and y), and then become 0 == z or 1 == z

Okay thank you

& and && is different

& is AND bitwise operator

Now days people ignoring compiler message and asking in the group instead

& is AND bitwise operator

I want AND operator. I found the issue. Thank you

I like how you ask to input 1, 2 and 3, compare them as x, y and z, and output to user as A, B and C 😆

I like how you ask to input 1, 2 and 3, compare them as x, y and z, and output to user as A, B and C 😆

Lol, i just realized it

What is :: scop resolution operator in c++

#res

eg log(123) = 2 * log(10) + 0 * log(2) + log(0.229)

Errm it is not. According to your equation log(123) = log(22.9) which it is definitely not. This equation is not correct. Either you have made a mistake somewhere or you misunderstood what they taught you at school.

Errm it is not. According to your equation log(123) = log(22.9) which it is definitely not. This equation is not correct. Either you have made a mistake somewhere or you misunderstood what they taught you at school.

x = 10^n * 2^m * d log(x) = n * log(10) + m * log(2) + log(d)

log(d) is found using a Taylor series expanded around x = 1 x = 10^n * 2^m * d log(x) = n * log(10) + m * log(2) + log(d) But base 10 is useless, we have numbers stored as order + mantissa, and we need to change the order keeping mantissa to get in the range from 0 to 1

result: log(x) = m * log(2) + log(x - 1)

What is :: scop resolution operator in c++

There are different kinds of scopes in C++ (scope is a region in which a name is valid). Ex - Namespace scope, class scope, block scope, global scope etc etc. The scope resolution operator is used to identify names within a class scope, a namespace scope, global scope or enumeration scope. For ex a global variable a can be identified using ::a. A name b in namespace scope N can be identified by N::b. A static name c in Class C can be identified as C::c. The name before :: identifies the scope. If it is empty it is assumed that you are referring to a name in global scope.

x = 10^n * 2^m * d log(x) = n * log(10) + m * log(2) + log(d)

This is not correct. Are you saying log(123) = 2*log(10) + 0*log(2) + log(0.229)? If this were true then the RHS is equal to log(22.9). Then log(123) would be equal to log(22.9). Doesn't this sound absurd to you?

This is not correct. Are you saying log(123) = 2*log(10) + 0*log(2) + log(0.229)? If this were true then the RHS is equal to log(22.9). Then log(123) would be equal to log(22.9). Doesn't this sound absurd to you?

Yeah, I was wrong, but the main thing is that everything is working now

I think it's the only proper way, why doesn't it suit you?

I guess it will be just if-elseif-else as in the example here https://en.cppreference.com/w/cpp/utility/any/any_cast There's no enum or similar thing underneath that would speed that up (because it can literally be any type, including any user-defined type).

You can also match by type_info, like in the example by the link below, its better when you have many possible variants https://en.cppreference.com/w/cpp/utility/any/type

But you anyway have to do any_cast in the end

what is the difference between <cstring> and <string.h>?

what is the difference between <cstring> and <string.h>?

just different style <string.h> is how you write it in valid C code <cstring> is how you use the same header file in C++ style to make it a bit different than C++ <string> library just not to be easily confused

Why is this not working? . . . char *ptr = "text""; *(ptr + 1) = 'o'; why do i get a segmentation fault ?

Why is this not working? . . . char *ptr = "text""; *(ptr + 1) = 'o'; why do i get a segmentation fault ?

you're writing to .rodata section, which is read only

you probably want this: char mut[] = "text"; char *ptr = mut; *(ptr + 1) = 'o';

that mut variable is mutable, so you can do a write dereference

but char *ptr = "text" is read only (the string literal "text" can't be changed), dereference only works for reading operation

Why is this not working? . . . char *ptr = "text""; *(ptr + 1) = 'o'; why do i get a segmentation fault ?

string literals are const char[] typed, thus any attempt to modify its element leads to undefined behavior. The implicit cast to char* should have been banned for type safety reason, but some compiler accept this with just a warning due to historical reason.

Why is this not working? . . . char *ptr = "text""; *(ptr + 1) = 'o'; why do i get a segmentation fault ?

the word text is stored somewhere called the “data segment” in the memory it only stores hard-coded values and they cannot change in runtime

string literals are const char[] typed, thus any attempt to modify its element leads to undefined behavior. The implicit cast to char* should have been banned for type safety reason, but some compiler accept this with just a warning due to historical reason.

neither gcc nor clang shows warning when compiling C code using: char * variable = “string” ;

the word text is stored somewhere called the “data segment” in the memory it only stores hard-coded values and they cannot change in runtime

string literals are typically stored in read-only memory section, but it's not guaranteed.

neither gcc nor clang shows warning when compiling C code using: char * variable = “string” ;

Of course no warning in C because the semantic difference in keyword const

I'm talking about C++ behavior, where the "historical problem" is exactly from C.

this hurts as a C programmer 😂

Hi guys, I have a shared memory question: so, Imagine you have a Process that creates A) A vector of N shared memoryes; B) N threads; Now, since the threads share the generator Process address space, I think it's safe to assume that, if I attach (with shmat) all the N shared memories in the code of the generator process, of course storing the address, then all the threads will be able to use this address in order to read/write information from/on it, right? Because, right now, I have something like this: PROCESS FUNCTION: Creates N-entry-shared memory vector with shmget (so the entry will be like: shmID0| shmID1 | ... | shmIDN-1); Attaches its address space only to the first one (for reasons due to the code itself); generates N threads; THREAD FUNCTION: Attaches itself on the shared memory he is interested in; But this is equivalent to the first thing I said, right? Even worst, if I attach the first shared memory (between generator process and first thread) both in process code and thread code, I would be attaching this thing twice, right? Thank you! Note: for those interested, I solved the problem and the solution is: it makes no sense to use shared memory with threads, just map k*N pages on the generators address space and pass the address to the page to the threads

you're writing to .rodata section, which is read only

But last time you said string literal is read-only and could live anywhere not only .rodata

But last time you said string literal is read-only and could live anywhere not only .rodata

depends on the context, when and where did i say that? let's take a look on the conversation

depends on the context, when and where did i say that? let's take a look on the conversation

https://t.me/programminginc/444243

It it lived on .rodata ? 🤔

it's not the const char * that places the data, but char *p assigned with a string literal const char *p doesn't always mean it points to .rodata, it can live everywhere const char *p = malloc(100) is fine too

https://t.me/programminginc/444243

string literal indeed can live everywhere

void fx() { char x[] = "hello"; } this lives on the stack

but char *p = "hello"; is immutable, that's why it's read only

But last time you said string literal is read-only and could live anywhere not only .rodata

i didn't say that

anyone?

the code is horrible and messed up, you have random indentations all over the place please fix that first before asking people to review your code, at least make it more human readable don't expect to get help if you don't help us to even properly read your code

https://linkode.org/#8fANgCGZAGRvIVmJIFULc6

the code is horrible and messed up, you have random indentations all over the place please fix that first before asking people to review your code, at least make it more human readable don't expect to get help if you don't help us to even properly read your code

ahh i see..thanks bro and sorry for the whole messed up code 🙏

https://linkode.org/#8fANgCGZAGRvIVmJIFULc6

Why are you calling pthread_create and pthread_join again in the inner loop? You don't need two loops here. You just need the outer loop. Printing to the console from different threads requires synchronisation between the threads. If you don't synchronize, the outputs will all be interleaved. So use something like a mutex to ensure that only one thread is writing to the console at a time. In the task method, the thread_num variable has a value between 1 and 5. Inside this method, lock the mutex and then use a loop to print as many stars as is the value of thread_num variable. Unlock the mutex after printing the stars.

Why are you calling pthread_create and pthread_join again in the inner loop? You don't need two loops here. You just need the outer loop. Printing to the console from different threads requires synchronisation between the threads. If you don't synchronize, the outputs will all be interleaved. So use something like a mutex to ensure that only one thread is writing to the console at a time. In the task method, the thread_num variable has a value between 1 and 5. Inside this method, lock the mutex and then use a loop to print as many stars as is the value of thread_num variable. Unlock the mutex after printing the stars.

I don't think calling pthread_join() right after pthread_create() is right. He still needs two loops, but not nested. The thread created by pthread_create() in this case is useless, he is only running a single thread at a time. The main thread will be blocking at pthread_join() right after the thread creation, waiting for the sub-thread get terminated. There is no parallelism gain. He totally misses the point of using pthread. You may want something like this: for (i = 0; i < n; i++) pthread_create(&p[i], ...); for (i = 0; i < n; i++) pthread_join(p[i], ...);

I don't think calling pthread_join() right after pthread_create() is right. He still needs two loops, but not nested. The thread created by pthread_create() in this case is useless, he is only running a single thread at a time. The main thread will be blocking at pthread_join() right after the thread creation, waiting for the sub-thread get terminated. There is no parallelism gain. He totally misses the point of using pthread. You may want something like this: for (i = 0; i < n; i++) pthread_create(&p[i], ...); for (i = 0; i < n; i++) pthread_join(p[i], ...);

Yeah. I missed that. He/she should call pthread_join after main creates all the threads.

I don't think calling pthread_join() right after pthread_create() is right. He still needs two loops, but not nested. The thread created by pthread_create() in this case is useless, he is only running a single thread at a time. The main thread will be blocking at pthread_join() right after the thread creation, waiting for the sub-thread get terminated. There is no parallelism gain. He totally misses the point of using pthread. You may want something like this: for (i = 0; i < n; i++) pthread_create(&p[i], ...); for (i = 0; i < n; i++) pthread_join(p[i], ...);

Yeah, that's right. If it were intentionally written like that, then you also miss the point of using mutex, because there is only single thread that actually runs.

There are different kinds of scopes in C++ (scope is a region in which a name is valid). Ex - Namespace scope, class scope, block scope, global scope etc etc. The scope resolution operator is used to identify names within a class scope, a namespace scope, global scope or enumeration scope. For ex a global variable a can be identified using ::a. A name b in namespace scope N can be identified by N::b. A static name c in Class C can be identified as C::c. The name before :: identifies the scope. If it is empty it is assumed that you are referring to a name in global scope.

How can use in class

How can use in class

You can use the scope operator in classes to access typedefs/aliases, inner classes, static members et al.

string choice = "y"; while (choice.compare("y") == 0) { something something} cout << "\nPlay again (y/n): "; getline(cin, choice); can someone explain to me how exactly is string.compare is being used here? to break out of the loop, it has to be false which means not equal to 0, but how does cin y make the loop be 0 (and continues looping again) ?

How to make a calculator which can perform operations more than 2 numbers? Eg: 2+7*7-8 If anybody have example project Please share

How to make a calculator which can perform operations more than 2 numbers? Eg: 2+7*7-8 If anybody have example project Please share

Like you want to code parser which would calculate these equations?

Like you want to code parser which would calculate these equations?

Yes, Maybe I will understand if I could see just concept.

And I guess you want to have operation priorities as well? Multiplication goes before addition etc?

And I guess you want to have operation priorities as well? Multiplication goes before addition etc?

Indeed😁

Then you will need to parse by these priorities eg. Search for *, then parse left and right number and then add them together. You can pre parse this by splitting it into some structure of each element

Then you will need to parse by these priorities eg. Search for *, then parse left and right number and then add them together. You can pre parse this by splitting it into some structure of each element

I tried many times but I made a program to calculate operations of 2 numbers only and another program for 3 numbers only(can't perform 2 numbers). Can't make proper