[英]Why is int x{ y = 5 } possible?
int main() {
int y;
int x{ y = 5 };
//x is 5
}
How is this possible, since y = 5 is not a calculable expression?这怎么可能,因为 y = 5 不是可计算的表达式?
Also, why doesn't the compiler or IDE complain about main() not returning an int?另外,为什么编译器或 IDE 不抱怨 main() 没有返回 int?
How is this possible, since y = 5 is not a calculable expression?
这怎么可能,因为 y = 5 不是可计算的表达式?
It is an assignment, and assignments yield values, ie the "cv-unqualified type of the left operand", see [expr.ass/3] .它是一个赋值,赋值产生值,即“左操作数的 cv 非限定类型”,参见[expr.ass/3] 。 Hence
y = 5
results in y
, which is 5
, which is used to initialize x
.因此
y = 5
导致y
,即5
,用于初始化x
。
With respect to your second question, see cppreference on main (or [basic.start.main/5] ):关于您的第二个问题,请参阅 cppreference on main (或[basic.start.main/5] ):
The body of the main function does not need to contain the
return
statement: if control reaches the end of main without encountering areturn
statement, the effect is that of executingreturn 0;
main function 的主体不需要包含
return
语句:如果控制到达 main 的末尾而没有遇到return
语句,则效果为执行return 0;
..
Hence, compiler or IDE warning you about a missing return
statement at the end of main
would be plain wrong.因此,编译器或 IDE 在
main
末尾警告您缺少return
语句是完全错误的。 Admittedly, the fact that you should always return
objects from non- void
functions execpt main
is kind of... well, for historical reason I guess.诚然,您应该始终从非
void
函数( execpt main
) return
对象的事实有点……好吧,我猜是出于历史原因。
The operator=()
results in a value, which is the value assigned to the variable. operator=()
产生一个值,该值是分配给变量的值。 Because of this, it is possible to chain assignments like this:因此,可以像这样链接分配:
int x, y, z;
x = y = z = 1;
I will start from your last question我将从你的最后一个问题开始
Also, why doesn't the compiler or IDE complain about main() not returning an int?
另外,为什么编译器或 IDE 不抱怨 main() 没有返回 int?
According to the C++ Standard (6.6.1 main function)根据C++标准(6.6.1主要功能)
5 A return statement in main has the effect of leaving the main function (destroying any objects with automatic storage duration) and calling std::exit with the return value as the argument.
5 main 中的 return 语句具有离开 main function(销毁具有自动存储持续时间的任何对象)并以返回值作为参数调用 std::exit 的效果。 If control flows off the end of the compound-statement of main, the effect is equivalent to a return with operand 0 (see also 18.3).
如果控制从 main 的复合语句的末尾流出,则效果等同于操作数为 0 的返回(另见 18.3)。
And relative to this question相对于这个问题
How is this possible, since y = 5 is not a calculable expression?
这怎么可能,因为 y = 5 不是可计算的表达式?
From the C++ Standard (8.18 Assignment and compound assignment operators)来自 C++ 标准(8.18 赋值和复合赋值运算符)
1 The assignment operator (=) and the compound assignment operators all group right-to-left.
1 赋值运算符 (=) 和复合赋值运算符都从右到左分组。 All require a modifiable lvalue as their left operand and return an lvalue referring to the left operand.
所有这些都需要一个可修改的左值作为它们的左操作数,并返回一个指向左操作数的左值。
Sp this declaration sp 这个声明
int x{ y = 5 };
can be equivalently split into two statements可以等价地分成两个语句
y = 5;
int x{ y };
Moreover in C++ you can even to make a reference to the variable y the following way此外,在 C++ 中,您甚至可以通过以下方式引用变量 y
int &x{ y = 5 };
Here is a demonstrative program这是一个演示程序
#include <iostream>
int main()
{
int y;
int &x{ y = 5 };
std::cout << "y = " << y << '\n';
x = 10;
std::cout << "y = " << y << '\n';
}
Its output is它的 output 是
y = 5
y = 10
You may this declaration你可以这个声明
int x{ y = 5 };
rewrite also like重写也喜欢
int x = { y = 5 };
However take into account that there is a difference between these (looking similarly as the above declarations) two declarations.但是请考虑到这两个声明(看起来与上述声明相似)之间存在差异。
auto x{ y = 5 };
and和
auto x = { y = 5 };
In the first declaration the variable x
has the type int
.在第一个声明中,变量
x
的类型为int
。 In the second declaration the variable x
has the type std::initializer_list<int>
.在第二个声明中,变量
x
的类型为std::initializer_list<int>
。
To make the difference more visible see how the values of the objects are outputted.为了使差异更加明显,请查看对象的值是如何输出的。
#include <iostream>
int main()
{
int y;
auto x1 { y = 5 };
std::cout << "x1 = " << x1 << '\n';
auto x2 = { y = 10 };
std::cout << "*x2.begin()= " << *x2.begin() << '\n';
std::cout << "y = " << y << '\n';
return 0;
}
The program output is程序 output 是
x1 = 5
*x2.begin()= 10
y = 10
If you take a look at the documentation on cppreference , you'll see that operator=()
return a reference to the object that was assigned.如果您查看 有关 cppreference 的文档,您会看到
operator=()
返回对已分配的 object 的引用。 Therefore, a assignment can be used as an expression that returns the object that was assigned.因此,赋值可以用作返回被赋值的 object 的表达式。
Then, it's just a normal assignment with braces.然后,这只是一个带大括号的正常赋值。
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