CODEWITHSUNDEEP
c++
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Reputation: 13

C++ error: no match for call to '(std::list<int>) (int&)'

I have a piece of script to count the highest possible sum of list elements while retaining the pattern: higher-lower-higher-lower-higher etc.

#include <algorithm>
#include <stdio.h>
#include <list>

using namespace std;

int main() {
    int n=0;
    bool g = false;
    int d = 1000000;
    list<int> a(d,0);
    int b;

    for (int x=0; x<n; x++) {  //load the list elements
        scanf("%d", &b);
        a.push_back(b);
    }

    for (int i=0; a(n) == a(n+1); i++) {
        a.remove(n+1);  //this is to remove any (consecutive) duplicates
    }


    if (a(n) > a(n+1)) {   
        g = false;
    } else {
        g = true;
    }


while (n+1 < d) {
    if (g) {
        if (!(a(n) < a(n+1))) {
            a.remove(n+1);
            continue;
        } else {
            n = n+1;
            g = !g;
            continue;
        }
    } else {
        if (!(a(n) > a(n+1))) {
            a.remove(n+1);
        } else {
            n = n+1;
            g = !g;
            continue;
        }
    }
}
long long sum = 0 ;
for (int i = 0 ; i < a.count(); i++) {
   sum += a(i);
}

printf("%lld ", sum);

return 0;
}

However, at compilation attempt it throws no match for call to '(std::list<int>) (int&)' everywhere where I am trying to compare list elements.

What am I doing wrong?

Upvotes: 0

Views: 2219

Answers (2)

Muhammad Nizami
Muhammad Nizami

Reputation: 926

a(n) means calling the function call operator std::list::operator()(int&) with n as its argument. The error no match for call to '(std::list<int>) (int&)' is because std::list doesn't have a call operator.

A list does not support random access unlike a vector which have operator[] for random access. Using a vector, the random access element n of vector a becomes a[n].

To traverse through a list, you use list iterators. Use std::list::begin to get the iterator to first element. To traverse, use assignment, increment (++) and decrement (--) operators. To access the value of the element, use dereference operator (*).

Upvotes: 0

Miles Budnek
Miles Budnek

Reputation: 30879

a(n) is not valid. std::list does not have a call operator (or a subscript operator, which is what I assume you intended to use). std::list does not provide random access.

If you need random access, use a container that provides it, such as std::vector. Then you can access items using a[n] notation.

Upvotes: 3

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