[英]Forcing std::vector overload instead of int overload on list with one element
Consider the code below: 考虑下面的代码:
#include <iostream>
#include <vector>
void f(std::vector<int> v) {std::cout << __PRETTY_FUNCTION__ << std::endl;}
void f(int n) {std::cout << __PRETTY_FUNCTION__ << std::endl;}
int main()
{
f({42}); // the int overload is being picked up
}
I was a bit surprised to realize that in this case the int overload is being picked up, ie the output of the program is: 我很惊讶地意识到在这种情况下会接收到int重载,即程序的输出为:
void f(int) 无效f(int)
with the warning 带有警告
warning: braces around scalar initializer [-Wbraced-scalar-init] f({42}); 警告:标量初始值设定项[-Wbraced-scalar-init] f({42})括起来;
Of course this happens only when I pass a 1-element list as an argument, otherwise the std::vector
overload is being picked up. 当然,只有当我将一个1元素列表作为参数传递时,才会发生这种情况,否则将拾取std::vector
重载。
Why is {42}
treated like a scalar and not like a init-list? 为什么{42}
被视为标量而不是初始化列表? Is there any way of forcing the compiler to pick the std::vector
overload (without explicitly constructing std::vector<int>{42}
) even on 1-element lists? 有没有办法强迫编译器选择std::vector
重载(而没有显式构造std::vector<int>{42}
)甚至在1元素列表上?
PS: The std::vector
has an init-list constructor PS: std::vector
具有init-list构造函数
vector(std::initializer_list<T> init, const Allocator& alloc = Allocator());
see (7) from cppreference . 请参阅cppreference中的 (7)。
Braced initializer has no type, we can't say {42}
is an int
or std::initializer_list<int>
. 大括号初始化器没有类型,我们不能说{42}
是int
或std::initializer_list<int>
。 When it's used as an argument, special rules for overload resolution will be applied for overloaded function call. 当用作参数时, 重载解析的特殊规则将应用于重载的函数调用。
(emphasis mine) (强调我的)
- Otherwise, if the parameter type is not a class and the initializer list has one element , the implicit conversion sequence is the one required to convert the element to the parameter type 否则,如果参数类型不是类,并且初始值设定项列表包含一个元素 ,则隐式转换序列是将元素转换为参数类型所需的序列
{42}
has only one element with type int
, then it's exact match for the overload void f(int)
. {42}
仅具有一个类型为int
元素,则它与重载void f(int)
完全匹配。 While for void f(std::vector<int>)
a user-defined conversion is needed. 对于void f(std::vector<int>)
需要用户定义的转换。 So void f(int)
will be picked up here. 因此, void f(int)
将在此处拾取。
Is there any way of forcing the compiler to pick the
std::vector
overload (without explicitly constructingstd::vector<int>{42})
even on 1-element lists? 有没有办法强迫编译器选择std::vector
重载(而无需显式构造std::vector<int>{42})
甚至在1元素列表上?
As a wordaround, you can put additional braces to force the compiler construct a std::initializer_list<int>
and then pick up void f(std::vector<int>)
: 可以用一个大括号将编译器构造为std::initializer_list<int>
,然后选择void f(std::vector<int>)
:
f({{42}});
Forcing std::vector overload 强制std :: vector重载
int main()
{
f(std::vector<int>{42}); // the vector overload is being picked up now
}
Why isn't the vector(initializer_list)
constructor being picked up? 为什么不选择vector(initializer_list)
构造函数?
Assume that another header declares a void f(std::set<int> v)
. 假设另一个标头声明一个void f(std::set<int> v)
。
How would you like the compiler to react when faced with f({1})
: construct a vector
or construct a set
? 您希望编译器在面对f({1})
时如何作出反应:构造vector
或构造set
?
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