Template functions
- C++ course
- Intermediate
- Lesson 48/116
- 5 min
Learn to define and call template functions.
Template function call
A template function is called like a normal function. In max(3, 5) the compiler deduces T=int. You can also specify the type: max<double>(3.0, 5.0).
Example: template <typename T> T max(T a, T b) { return a > b ? a : b; }. Here a > b ? a : b is a ternary that returns the larger value. The call max(3, 5) yields 5. The compiler produces a separate version for each call; there is no run-time overhead.
Type deduction limits
Type deduction doesn't always work. In max(3, 5.0) the compiler cannot decide whether T is int or double; the two arguments have different types.
The fix is to specify the type: max<double>(3, 5.0) runs with T=double; 3 is converted to double. Likewise max<int>(3, 5.0) converts 5.0 to int. When deduction is ambiguous, an explicit type is required.
Practice Write the add template
Write template <typename T> T add(T a, T b) that returns a + b. In main call std::cout << add(3, 4) << std::endl;.
Starter code
#include <iostream>
// write the add template function: T add(T a, T b)
int main() {
// call add and print the result
return 0;
}Sample solution
#include <iostream>
template <typename T>
T add(T a, T b) {
return a + b;
}
int main() {
std::cout << add(3, 4) << std::endl;
return 0;
}Expected output
7
Questions and answers
Review the lesson questions together with their answers. To test yourself, solve the lesson interactively.
1 Match call to meaning
| Term | Meaning |
|---|---|
max(3, 5) |
T=int deduced |
max<double>(3.0, 5.0) |
Explicit type |
max("a", "b") |
T=const char* |
max<int>(3, 5) |
Explicit int |
2 What does T become in the call max(3, 5)?
template <typename T> T max(T a, T b);- double
- int Correct answer
- const char*
- void
Explanation: 3 and 5 are int literals, so T=int is deduced.
3 Complete the explicit-type call
int result = max____(3, 5);int result = max<int>(3, 5);Explanation: max<int>(3, 5) specifies the type as int explicitly.