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Kyrion
Kyrion

Reputation: 188

Zero-initialized, unnamed, temporary unsigned int

Am I right to interpret unsigned int{}; as 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

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{});
}

Upvotes: 4

Views: 119

Answers (2)

songyuanyao
songyuanyao

Reputation: 172894

Clang and Gcc are right. unsigned int{}; is considered as explicit type conversion firstly, which only works with single-word type name; while unsigned int (and int* etc) is not but int is.

(emphasis mine)

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.

As the workaround, you can 

using unsigned_int = unsigned int;
unsigned_int{};

As the result it'll value-initialize (zero-initialize) a temporary unsigned int as you expected.

Upvotes: 3

R Sahu
R Sahu

Reputation: 206577

You can use a type alias to get around the limitation.

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

Upvotes: 2

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