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Commit 78b1cdf

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book: correct description of function template deduction from auto (#280)
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‎book/en-us/02-usability.md‎

Lines changed: 11 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -418,17 +418,23 @@ auto i = 5; // i as int
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auto arr = new auto(10); // arr as int *
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```
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Since C++ 20, `auto` can even be used as function arguments. Consider
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the following example:
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Since C++ 14, `auto` can even be used as function arguments in generic lambda expressions,
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and such functionality is generalized to normal functions in C++ 20.
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Consider the following example:
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```cpp
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int add(auto x, auto y) {
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auto add14 = [](auto x, auto y) -> int {
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return x+y;
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}
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int add20(auto x, auto y) {
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return x+y;
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}
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auto i = 5; // type int
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auto j = 6; // type int
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std::cout << add(i, j) << std::endl;
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std::cout << add14(i, j) << std::endl;
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std::cout << add20(i, j) << std::endl;
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```
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> **Note**: `auto` cannot be used to derive array types yet:
@@ -481,7 +487,7 @@ type z == type x
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### tail type inference
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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++:
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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++:
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```cpp
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template<typename R, typename T, typename U>

‎book/zh-cn/02-usability.md‎

Lines changed: 11 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -354,17 +354,22 @@ auto i = 5; // i 被推导为 int
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auto arr = new auto(10); // arr 被推导为 int *
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```
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从 C++ 20 起,`auto` 甚至能用于函数传参,考虑下面的例子:
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从 C++ 14 起,`auto` 能用于 lambda 表达式中的函数传参,而 C++ 20 起该功能推广到了一般的函数。考虑下面的例子:
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```cpp
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intadd(auto x, auto y) {
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auto add14 = [](auto x, auto y) -> int {
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return x+y;
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}
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auto i = 5; // 被推导为 int
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auto j = 6; // 被推导为 int
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std::cout << add(i, j) << std::endl;
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int add20(auto x, auto y) {
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return x+y;
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}
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auto i = 5; // type int
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auto j = 6; // type int
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std::cout << add14(i, j) << std::endl;
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std::cout << add20(i, j) << std::endl;
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```
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>
@@ -413,7 +418,7 @@ type z == type x
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### 尾返回类型推导
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你可能会思考,在介绍 `auto` 时,我们已经提过 `auto` 不能用于函数形参进行类型推导,那么 `auto` 能不能用于推导函数的返回类型呢?还是考虑一个加法函数的例子,在传统 C++ 中我们必须这么写:
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你可能会思考, `auto` 能不能用于推导函数的返回类型呢?还是考虑一个加法函数的例子,在传统 C++ 中我们必须这么写:
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```cpp
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template<typename R, typename T, typename U>

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