This article explains how to create more generic functions using type deduction to increase code maintainability in C++ programming. Type deduction is the process by which the compiler automatically deduces the types of variables and parameters used in function calls. For example, if a function accepts an int parameter and you pass a double, then the compiler will infer the type of the variable from the parameter. This article explains the process of type deduction and shows how to use it to create more generic functions.
Type deduction in C++ is a powerful technique that can be used to increase code maintainability and help prevent mistakes. To achieve this, C++ compilers use type inference to deduce types and eliminate type-related errors in the code.
However, type inference can have its limits. Sometimes, it is hard to determine the type of variables and arguments passed into a function. In such cases, it can be useful to add additional constraints to the code, or add a hint for the compiler to deduce the correct type. The following example shows how to create more generic functions using type deduction to increase code maintainability in C++ programming.
To see how type deduction works in practice, consider the following code:
void foo(int x) { }
int main() {
int x = 2;
foo(x); // error, `foo` expects an integer
}
When this code is compiled, the compiler will generate an error:
error: conversion from ‘double’ to non-scalar type ‘int’ requested
The compiler cannot deduce the type of x, so it throws the above error. However, if we specify that x is of type int, then the code compiles without any errors.
To achieve this, we need to specify the type of the variable in the call to foo().
In the above example, the compiler will deduce the type of x based on the type of x in the body of the function. Therefore, to make the compiler deduce the type of x, we need to add an explicit constraint to the function. To achieve this, we need to specify the type of x as follows:
void foo(int x) { }
int main() {
int x = 2;
foo(x); // ok, compiler can deduce the type of x
}
In the above code, the compiler will infer the type of x as int, since it is an int in the main() function. Therefore, the compiler will not throw an error, and the code will compile successfully.
As shown in the above example, the compiler can deduce the type of the parameter x only when the function body does not explicitly use the type of the parameter. In the previous example, the function body does not use the type of x, and therefore the compiler cannot deduce the type of x. However, we can still specify the type of x in the call to foo(), and the compiler will deduce the type of x.
There are some cases where type deduction does not work. For example, consider the following code:
void foo(int x, int y) { }
int main() {
int x = 2;
foo(x, 3); // error, compiler does not deduce the type of `y`
}
As shown in the above example, the compiler will generate an error. This happens because the compiler cannot deduce the type of x from the type of x in the body of the function. Therefore, the compiler cannot infer the type of x and y.
If we want to enable the compiler to deduce the type of x and y, we need to add a type constraint to the function.
void foo(int x, int y) { }
int main() {
int x = 2;
foo(x, 3); // ok, compiler will deduce the type of x and y
}
In the above code, the compiler will deduce the type of x as int, and the type of y as int. Therefore, the compiler will not throw an error, and the code will compile successfully.
The compiler can deduce the type of the parameters only when the function body uses the type of the parameters. In the previous example, the function body uses the type of x, but does not use the type of y. Therefore, the compiler cannot deduce the type of x and y. However, we can still specify the types of x and y in the call to foo().
Although type deduction can sometimes be hard to use, it has several advantages. For example, it helps to reduce the amount of code, because we no longer need to explicitly specify the type of the variables and parameters in the function call. It also helps to prevent mistakes, because the compiler will check that the types of the variables and parameters in the function call match the types of the variables and parameters in the function body. It also helps to increase the readability of the code, because the compiler will automatically insert type declarations for variables and parameters in the function body, and it will automatically infer the type of the variables and parameters in the function body.
C++ allows you to make use of type deduction while working with functions by with two different ways. Types can be deduced from parameters of functions by creating a template and then calling it without any explicit specialist operators. The type of return can be determined for a function by with the keyword auto in lieu of the return type. Below program illustrates the use of auto to discover the return type of an operation.
#include
using namespace std;
auto AutoFunctionFromReturn(int parameter)
{
return parameter;
}
int main()
{
auto value = AutoFunctionFromReturn(1);
cout << value << endl;
return 0;
}
The AutoFunctionFromReturn function’s return type in the above program is automatically deduced. The compiler analyzes the nature of the variable returned by the function, and then utilizes this information to calculate the type that will be returned. All of this is done correctly because the compiler is equipped with all the information it needs within the function to determine the nature of the variable. It is the parameter variable that’s returned, so the compiler is able to make use of it as the type of return of the function.
#include
using namespace std;
template
auto AutoFunctionFromParameter(T parameter) -> decltype(parameter)
{
return parameter;
}
int main()
{
auto value = AutoFunctionFromParameter(2);
cout << value << endl;
return 0;
}
The auto function to work in functions that do not have any trailing returns type when it is used in conjunction with templates function
#include
using namespace std;
template
auto AutoFunctionFromParameter(T parameter)
{
return parameter;
}
int main()
{
auto value = AutoFunctionFromParameter(2);
cout << value << endl;
return 0;
}
