Please which compiler are you guys using

oh and clang is more descriptive

Hello, I am running a large data set in a for loop that exceeds the maximum number of array. (Over 10^8 integers). Does anybody knows any approach to store my data into smaller sizes to optimize memory.

Suppose the following values are inserted into a binary tree, in the order given: 12,7,9,10,22,24,30,18,3,14,20 Draw a diagram of the resulting binary tree

gcc

What

Hello, I am running a large data set in a for loop that exceeds the maximum number of array. (Over 10^8 integers). Does anybody knows any approach to store my data into smaller sizes to optimize memory.

Recursion?

Recursion requires more memory than iterating

Hello, I am running a large data set in a for loop that exceeds the maximum number of array. (Over 10^8 integers). Does anybody knows any approach to store my data into smaller sizes to optimize memory.

while(foo.currrent()){ foo.exec()}

save to file and read back in chunks?

Recursion requires more memory than iterating

#include <stdio.h> #include <stdlib.h> struct frame { int value; int count; }; int getlfu(struct frame *f, int n) { int max = 0; for (int i = 0; i < n; i++) { if (f[i].value == -1) return i; if (f[max].count < f[i].count) { max = i; } } return max; } int main() { int i, j, nop, n, hit = 0, fault = 0, pos = 0, flag = 0, choice, x = 1000; printf("Enter the no of frames in main memory:"); scanf("%d", &n); printf("Enter the no of values in reference string"); scanf("%d", &nop); int page[nop]; struct frame f[n]; printf("Enter the refernce string:"); for (i = 0; i < nop; i++) { scanf("%d", &page[i]); if (i < n) { f[i].value = -1; f[i].count = 0; } } for (i = 0; i < n; i++) { f[i].value = -1; f[i].count = 0; } for(i=0;i<nop;i++) printf("%d ",page[i]); fault=hit=0; for (i = 0; i < nop; i++) { for (j = 0; j < n; j++) { printf("page[%d] is %d and f[%d].value is %d\n",i,page[i],j,f[j].value); if (page[i] == f[j].value) { flag = 1; hit++; f[j].count=0; for (int k = i + 1; k < nop; k++) { if (f[j].value == page[k]) break; else f[j].count++; } for (int k = 0; k < nop && f[k].value != -1; k++) { if (k != j) f[k].count--; } break; } } if (flag == 0) { pos = getlfu(f, n); f[pos].value = page[i]; fault++; f[pos].count = 0; for (int k = i + 1; k < nop; k++) { if (f[pos].value == page[k]) break; else f[pos].count++; if (k == nop - 1) f[pos].count = --x; } for (j = 0; j < n && f[j].value != -1; j++) { if (j != pos) f[j].count--; } } printf("\n"); flag = 0; for (j = 0; j < n; j++) { if (f[j].value != -1) printf("%d ", f[j].value); else printf("- "); } printf("\n"); } printf("fault:%d hit:%d hit ratio:%d/%d\n", fault, hit, hit, hit + fault); for(i=0;i<nop;i++) printf("%d ",page[i]); }

Could anyone help me with this. The value of page[i] is changing at the end but i didn't even modify it. Some help please

Sorry Guys am a beginner in C++ can any of you help this Write a CPP program to implement binary tree.

Sorry Guys am a beginner in C++ can any of you help this Write a CPP program to implement binary tree.

if you want to be that lazy and cheat, just ask google...

This is really easy to find in google, what problems you have with the implementation?

Please help guys.... I'm interested in printing the output on an excel document..... But after running the program.... I found nothing happened when opening the excel document......

printing the output into a excel document?

The problem with me is that long term code during the implimantation of binary tree, I therefore humbly here for anyone with simple idea

What kind of help you need? Have you read the rules, by the way?

I'm afraid to write here because there are a lot of rules 😐

Well, the most gereric thing is don't ask to do your work for you (homework, test, etc.). Asking good questions is difficult, but learning this can be very beneficial (really, so many people don't get good answers in these chats because they asked their questions poorly). Of course, don't do some generally not nice things like: sending adverts, non-english, nsfw, off topic or spaming, I think this should be logical. Don't ask for, or propose pirated content. Other stuff in the rules is just specific repeatedly happening cases that people got annoyed with like questions "hack Instagram account for me plz".

Oh, and that sort of things. I understood, thx

And about the homework stuff, it's still fine to ask specific questions like "I need to do A, to do that I did B, I expected to get C as a result, but get D. Why?" even if this homework this is fine because A) you demonstrated that you made effort to solve it yourself and B) asked specific question about the problem you want to solve (with describing some useful information about the problem: what you tried, expected, got)

Hi everyone! I'm trying to detect which C standard is defined to run my code in header file. According to IMB on standard predefined macro-names: "__STDC_VERSION__ (C only) The integer constant of type long int: 199409L for the C89 language level, 199901L for C99. Note that this macro is only defined if __STDC__ is also defined." But so far it yields no result. What am I doing wrong here? header.h #define <stdio.h> #if defined(__STDC__) #if STDC_VERSION >= 199901L #define OPTION 1 #endif #if STDC_VERSION == 199409L #define OPTION 2 #endif #endif main.c #include "header.h" int main(void) { printf("%i\n", OPTION); return (0); }

I assume that header.h uses #include <stdio.h> instead of #define <stdio.h> What compiler are you using? If it is gcc or Clang, are you compiling using -traditional option.

Yes, it’s a typo near the stdio. I’m running it with gcc -std=c89 main.c or gcc -std=c99 main.c

Try this #if defined (_ _STDC_ _) #if defined(_ _STDC_VERSION_ _) && (_ _STDC_VERSION_ _ >= 199901L) #define OPTION 1 #else #define OPTION 2 #endif #endif I have left gaps in the double underscores because Telegram would format it as italicized text otherwise

Hello can you help me in opengl

Hey, guys! I am trying to calculate some expression using asin() function in C. When I try to print it like this: printf("expr = %lf", result); It prints expr = nan Why not a number printing, can asin() from <math.h> return "nan"? I checked all my expression, but it still returns nan. Here it is: double result = asin((i - a.y) / radius) + (grad * M_PI / 180.0); radius is double a.y is int grad is double i is int Radius is not a 0. With some values a.y and radius I have result nan, with other - a basic number result. What am I doing wrong?

As I see, argument of asin should be in range [-1.0; 1.0] otherwise it's an error, and NaN can be returned. https://en.cppreference.com/w/cpp/numeric/math/asin Can you check that the arguments are in the range (print them also to see)

As I see, argument of asin should be in range [-1.0; 1.0] otherwise it's an error, and NaN can be returned. https://en.cppreference.com/w/cpp/numeric/math/asin Can you check that the arguments are in the range (print them also to see)

It would be better to ask the whole question with the context.

#include<iostream> using namespace std; int main() { int arr[5]; for (int i{}; i < 5; i++) cin >> arr[i]; //work here auto initilaize = [arr] { for (int i{}; i < 5; i++) cin>>arr[i]; //don't work here cout <<'\n'; }; initilaize(); return 0; }