零初始化、未命名、临时 unsigned int

Question

我将 unsigned int{}; 解释为 zero-initialization of a temporary unsigned int without a name in the contexts provided below? The example does not work with clang or gcc, but compiles fine in Visual Studio: https://godbolt.org/z/LYWCCN

对吗

int ifun(int value) {
    return value * 10;
}

unsigned int uifun(unsigned int value) {
    return value * 10u;
}

int main() {
    // does not work in gcc and clang
    unsigned int{};
    unsigned int ui = uifun(unsigned int{});

    // works with all three compilers, also unsigned{}; works
    int{};
    int i = ifun(int{});
}

Answer 1

Clang 和 Gcc 是对的。 unsigned int{}; 首先被认为是 explicit type conversion，它只适用于单字类型名称；虽然 unsigned int（和 int* 等）不是，但 int 是。

（强调我的）

5) A single-word type name followed by a braced-init-list is a prvalue of the specified type designating a temporary (until C++17) whose result object is (since C++17) direct-list-initialized with the specified braced-init-list.

作为解决方法，您可以

using unsigned_int = unsigned int;
unsigned_int{};

作为结果，它将按您的预期值初始化（零初始化）一个临时 unsigned int。

Answer 2

您可以使用类型别名来绕过限制。

using UI = unsigned int;
UI{};
UI ui = uifun(UI{});

零初始化、未命名、临时 unsigned int

Zero-initialized, unnamed, temporary unsigned int

c++

gcc

initialization

clang

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