未执行左值到右值的转换

Question

下面的函数returns an rvalue:

int foo()
{
    int x = 42;
    return x;    // x is converted to prvalue
}

Clang 的 AST also shows the conversion:

`-FunctionDecl <line:1:1, line:5:1> line:1:5 foo 'int ()'
  `-CompoundStmt <line:2:1, line:5:1>
    |-DeclStmt <line:3:5, col:15>
    | `-VarDecl <col:5, col:13> col:9 used x 'int' cinit
    |   `-IntegerLiteral <col:13> 'int' 42
    `-ReturnStmt <line:4:5, col:12>
      `-ImplicitCastExpr <col:12> 'int' <LValueToRValue>
                                         ^^^^^^^^^^^^^^
        `-DeclRefExpr <col:12> 'int' lvalue Var 0x627a6e0 'x' 'int'

下面还执行左值到右值的转换，这次是针对进入函数的参数。

void f(int i) {}
int main()
{
    int x{3};
    f(x);
}

AST includes the conversion:

`-FunctionDecl <line:2:1, line:6:1> line:2:5 main 'int ()'
  `-CompoundStmt <line:3:1, line:6:1>
    |-DeclStmt <line:4:5, col:13>
    | `-VarDecl <col:5, col:12> col:9 used x 'int' listinit
    |   `-InitListExpr <col:10, col:12> 'int'
    |     `-IntegerLiteral <col:11> 'int' 3
    `-CallExpr <line:5:5, col:8> 'void'
      |-ImplicitCastExpr <col:5> 'void (*)(int)' <FunctionToPointerDecay>
      | `-DeclRefExpr <col:5> 'void (int)' lvalue Function 0x6013660 'f' 'void (int)'
      `-ImplicitCastExpr <col:7> 'int' <LValueToRValue>
                                        ^^^^^^^^^^^^^^
        `-DeclRefExpr <col:7> 'int' lvalue Var 0x60138a0 'x' 'int'

按我的理解，同理，下面应该也需要左值到右值的转换。

struct A{};
void f(A a) {}
int main()
{
    A a;
    f(a);
}

但是 never shows up in the AST:

`-CallExpr <line:6:5, col:8> 'void'
  |-ImplicitCastExpr <col:5> 'void (*)(A)' <FunctionToPointerDecay>
  | `-DeclRefExpr <col:5> 'void (A)' lvalue Function 0x615e830 'f' 'void (A)'
  `-CXXConstructExpr <col:7> 'A' 'void (const A &) noexcept'
    `-ImplicitCastExpr <col:7> 'const A' lvalue <NoOp>
      `-DeclRefExpr <col:7> 'A' lvalue Var 0x615ea68 'a' 'A'

为什么？转换有时是可选的吗？

Answer 1

AST 显示解析了 A 的构造函数，即采用左值 const A&（main 中的 a）并构造 a 在 f(A a) 中。这里没有右值。

`-CallExpr <line:6:5, col:8> 'void'
  |-ImplicitCastExpr <col:5> 'void (*)(A)' <FunctionToPointerDecay>
  | `-DeclRefExpr <col:5> 'void (A)' lvalue Function 0x615e830 'f' 'void (A)'
  `-CXXConstructExpr <col:7> 'A' 'void (const A &) noexcept'
    ^^^^^^^^^^^^^^^^                    ^^^^^^^^^
    `-ImplicitCastExpr <col:7> 'const A' lvalue <NoOp>
      `-DeclRefExpr <col:7> 'A' lvalue Var 0x615ea68 'a' 'A'

Answer 2

Why? Is the conversion optional sometimes?

不需要，已屏蔽。

对于 class 类型 A，f(a); 导致调用 A 的复制构造函数。隐式定义的复制构造函数采用左值引用（即 const A&），并且在绑定左值引用时抑制左值到右值的转换。

[dcl.init.ref]/5.1:

(5.1) If the reference is an lvalue reference ...

...

[ Note: The usual lvalue-to-rvalue, array-to-pointer, and function-to-pointer standard conversions are not needed, and therefore are suppressed, when such direct bindings to lvalues are done. — end note ]

未执行左值到右值的转换

Lvalue to rvalue conversion not performed

c++

clang

lvalue-to-rvalue