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book: correct description of function template deduction from auto (#280)

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A. Jiang
2024-05-21 05:58:17 +08:00
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parent 5c50dbc591
commit 78b1cdf6bb
2 changed files with 22 additions and 11 deletions
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16
book/en-us/02-usability.md
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@@ -418,17 +418,23 @@ auto i = 5; // i as int
auto arr = new auto(10); // arr as int *
```
Since C++ 20, `auto` can even be used as function arguments. Consider
the following example:
Since C++ 14, `auto` can even be used as function arguments in generic lambda expressions,
and such functionality is generalized to normal functions in C++ 20.
Consider the following example:
```cpp
int add(auto x, auto y) {
auto add14 = [](auto x, auto y) -> int {
return x+y;
}
int add20(auto x, auto y) {
return x+y;
}
auto i = 5; // type int
auto j = 6; // type int
std::cout << add(i, j) << std::endl;
std::cout << add14(i, j) << std::endl;
std::cout << add20(i, j) << std::endl;
```
> **Note**: `auto` cannot be used to derive array types yet:
@@ -481,7 +487,7 @@ type z == type x
### tail type inference
You may think that when we introduce `auto`, we have already mentioned that `auto` cannot be used for function arguments for type derivation. Can `auto` be used to derive the return type of a function? Still consider an example of an add function, which we have to write in traditional C++:
You may think that whether `auto` can be used to deduce the return type of a function. Still consider an example of an add function, which we have to write in traditional C++:
```cpp
template<typename R, typename T, typename U>