Given an int variable x write some statements that attempt to open a file named "table20" and read a value into x; if that turns out not to be possible your code should then read avalue from standard input into x.

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Click on the title for the solution
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This is the answer:

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ifstream filename;
filename.open("table20");
if (filename.fail()){
    cin >> x;
}else{
    filename >> x;
}

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. . Click on the title for the solution . . This is the answer: : #include <iostream> #include <fstream> #include <string> using namespace std; void lookup(); void add(); int main() { string type; while (cin >> type) { if (type == "lookup") lookup(); else  add(); } return 0; } void lookup() { ifstream fin; fin.open("phonedir"); if(!fin.is_open()) cout << "Not Open" << endl; bool check = false; string name,firstName,lastName,email; cin >> name; while (fin >> firstName >> lastName >> email) { if (firstName== name || lastName==name) { cout<<firstName<<" "<<lastName<<" "<<email<<endl; check=true; } } if(!check) cout << name << ": not found" << endl; fin.close(); } void add() { ofstream fout; string name; string fir...
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