空值不会被忽略，因为它应该是错误

Question

I am using Doubly Linked List Data Structure of CPP and have written the following code

#include<iostream>
#include <bits/stdc++.h> 
using namespace std;


int main()
{
        
    
    list <int> l1[64]; 
    l1[1].push_back(100);
    l1[2].push_back(200);
    int p=l1[1].pop_front();
    cout<<p<<endl;
     p=l1[2].pop_front();
    cout<<p<<endl;
    

}

But I am getting this errors-->

trial.cpp: In function ‘int main()’:
trial.cpp:15:23: error: void value not ignored as it ought to be
   15 |  int p=l1[1].pop_front();
      |        ~~~~~~~~~~~~~~~^~
trial.cpp:17:20: error: void value not ignored as it ought to be
   17 |   p=l1[2].pop_front();
      |     ~~~~~~~~~~~~~~~^~

I am unable to understand the reason for this error, because I am just popping the first element into the p variable and printing it. Then why am I getting this error?

Answer 1

pop_front does not return anything. The reason is that it would be impossible to implement the strong exception safety guarantee if it did. To access the front element of your list use front . Here's your code rewritten.

int main()
{    
    list <int> l1[64]; 
    l1[1].push_back(100);
    l1[2].push_back(200);
    int p=l1[1].front();
    l1[1].pop_front();
    cout<<p<<endl;
    p=l1[2].front();
    l1[2].pop_front();
    cout<<p<<endl;
}