Expected primary-expression before ‘long’

Question

What is the problem with my code ?

prog.cpp: In function ‘void helper()’: prog.cpp:15:25: error: expected primary-expression before ‘long’ ans = ans + max(long long(0),vec[i]-i);

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007

void helper(){
    ll n;
    cin>>n;
    vector<ll>vec(n);
    for(ll i=0;i<n;i++)
        cin>>vec[i];
    sort(vec.begin(),vec.end(),greater<ll>());
    ll ans=0;
    for(ll i=0;i<n;i++){
        ans = ans + max(long long(0),vec[i]-i);
        ans=ans % mod;
    }
    cout<<ans<<endl;
}

int main() {
    int test;cin>>test;
    while(test--){
        helper();
    }
    
    return 0;
}

Use `max((long long)0, ven[i]-i);`, or `max( 0ll, vec[i]-i );` — Tim Roberts, Feb 04 '22 at 07:40
*Please* inform whatever fiend is telling you and your classmates to use `#include ` that they're doing you no favors and [it's a terrible habit](https://stackoverflow.com/questions/31816095/why-should-i-not-include-bits-stdc-h) — WhozCraig, Feb 04 '22 at 08:21

score 2 · Answer 1 · answered Feb 04 '22 at 08:13

2

You can correct the error following one of these ways: (ll)0 or (long long)0 or 0ll

answered Feb 04 '22 at 08:13

Md. Faisal Habib

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I followed these 3 steps also but it is not working , showing error – SUKRITI GUIN Feb 04 '22 at 09:05
1

I ran code properly using above approach. Change code line : ans = ans + max( 0ll,vec[i]-i); Can you please share what error you're getting after changing? – Md. Faisal Habib Feb 04 '22 at 09:09

score 0 · Answer 2 · answered Feb 04 '22 at 10:20

The formal rule is that a constructor like type cast like long long(0) can only have a one word type name, and long long is two words. Oops, just doesn't work.

Using a proper C++ type cast like static_cast<long long>(0) does work. However, for a constant you could just use 0LL and avoid the cast altogether.

Expected primary-expression before ‘long’

2 Answers2