Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 8136345
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: June 6, 20262026-06-06T10:44:01+00:00 2026-06-06T10:44:01+00:00

int lf = ((t.left==null) = (t.right==null)) ? 1:0; it returns 1 if the statement

  • 0
int lf = ((t.left==null) = (t.right==null)) ? 1:0;

it returns 1 if the statement in the bigger parenthesis is true, but in the middle, whats the point of assigning right value to lefT?

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-06-06T10:44:03+00:00Added an answer on June 6, 2026 at 10:44 am

    Normally you’d have an equal sign to assign. The return of the assigned is the same as the RHS of the expression.

    You’d use an equal sign in a expression within an if to assign and check the result at the same time.

    // return first and third items added if they exist.
    if ((list = GetItems()).Length > 2) { return list[0] + list[2]; } 
    

    Right here all you have is a compiler error because t.left==null evaluates to (true/false) and you can’t assign to that.

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

#ifndef _BST_H_ /* Returns negative (left<right), zero (left==right), or positive (left>right). */ typedef int
I'm running code that sometimes yields this: UInt32 current; int left, right; ... //sometimes
typedef struct nodetype { int data; struct nodetype * left; struct nodetype * right;
On left shift of (char) 0xff by 8 and casting it to int we
Here's the node definition: struct node{ int data; stuct node * left; struct node
struct node* insert(struct node* node, int data) { if(node == NULL) { // if
Ok this looks better, but gives a null pointer exception ? public static BST
Having construction in a form: struct Node { Node():left_(nullptr), right_(nullptr) { } int id_;
int somefunction(bool a) { try { if(a) throw Error(msg); return 2; } catch (Error
int main() { int var = 0;; // Typo which compiles just fine }

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.