How are types therein user-defined conversion?

1.	How are types therein user-defined conversion?
A.	4
B.	3
C.	2
D.	1
E.	None of these
Answer» D. 1

Discussion

No Comment Found

Related MCQs

What is the output of this program? <pre class="prettyprint lang-c">#include <iostream> using namespace std; int main() { int p; p = 15 + 13 * 6; cout << p; return 0; } </iostream></pre>
What is the output of this program? <pre class="prettyprint lang-c">#include <iostream> using namespace std; int main() { int p = 7, q = 9, r; r = (p > q) ? p : q; cout << r; return 0; } </iostream></pre>
What is the output of this program? <pre class="prettyprint lang-c">#include <iostream> using namespace std; int main () { int p, q; p = 6; q = ++p * ++p; cout << p q = p++ * ++p; cout < return 0; } </iostream></pre>
What is the output of this program? <pre class="prettyprint lang-c">#include <iostream> using namespace std; class Example { public: int m, n; Example() {}; Example(int, int); Example operator + (Example); }; Example::Example (int p, int q) { m = p; n = q; } Example Example::operator+ (Example parameter) { Example container; container.m = m + parameter.m; container.n = n + parameter.n; return (container); } int main () { Example p (3,4); Example q (6,5); Example R; R = p + q; cout << R.m << "," << R.n; return 0; } </iostream></pre>
What is the output of this program? <pre class="prettyprint lang-c">#include <iostream> using namespace std; void Add(int p, int q, int & r) { p = q + r; p = q + r; r = p + q; } int main() { int n = 6, m =4; Add(n, m, m); cout << n << " " << m; return 0; } </iostream></pre>
What is the output of this program? <pre class="prettyprint lang-c">#include <iostream> using namespace std; ostream & operator< { return k; } int main() { cout << 10 << endl; cin.get(); return 0; } </iostream></pre>
What is the output of this program? <pre class="prettyprint lang-c">#include <iostream> #include <string> using namespace std; class Example { public: operator string () { return "Converted..."; } }; int main() { Example examp; string str = examp; cout << str << endl; return 0; } </pre>
What is the output of this program? <pre class="prettyprint lang-c">#include <iostream> #include <cmath> using namespace std; class ComplexNumber { private: double RealNumber; double ImagNumber; public: ComplexNumber(double r = 0.0, double i = 0.0) : RealNumber(r), ImagNumber(i) {} double mag() { return getMag(); } operator double () { return getMag(); } private: double getMag() { return sqrt(RealNumber * RealNumber + ImagNumber * ImagNumber); } }; int main() { ComplexNumber comp(2.0, 5.0); cout << comp.mag()< cout << comp; return 0; } </pre>
What is the output of this program? <pre class="prettyprint lang-c">#include <iostream> using namespace std; class Example { public: Example(int k) : num_k(k) { } public: int operator()(int k = 0) const { return num_k + k; } operator int () const { return num_k; } private: int num_k; friend int g(const Example&); }; int fun(char Z) { return Z; } int main() { Example fun(6); cout << fun(5); return 0; } </pre>
What is the output of this program? <pre class="prettyprint lang-c">#include <iostream> using namespace std; main() { double num = 23.9399; float num0 = 20.23; int p ,q; p = (int) num; q = (int) num0; cout < } </iostream></pre>

How are types therein user-defined conversion?

Discussion

No Comment Found

Related MCQs

Reply to Comment