什么是C ++ lambda函数的默认调用约定? [英] What's the default calling convention of a C++ lambda function?
问题描述
下面的代码是用VC ++ 2012编译的:
void f1(void(__stdcall *)())
{}
$ b $ void f2(void(__cdecl *)())
{}
void __cdecl h1()
{}
void __stdcall h2()
{}
int main()
{
f1(h1); //错误C2664
f2(h2); // error C2664
f1([](){}); // OK
f2([](){}); // OK
auto fn = [](){};
f1(fn); // OK
f2(fn); // OK
我认为错误是正常的,但是OK是不正常的。 / p>
所以,我的问题是:
-
如何指定C ++ lambda函数的调用约定?
-
编译器是否自动生成多个版本的lambda函数?即如下面的伪代码:
[] __stdcall(){};
[] __cdecl( ){};在VC ++ 2012上,编译器选择自动调用无状态的转换。
lambda表达式(没有捕获变量)当你将无状态lambda转换为函数指针。 What's the calling convention of a C++ lambda function?
How to specify the calling convention of a C++ lambda function?
If the calling convention is not defined, how to correctly recycle the stack space after having called a lambda function?
Does the compiler automatically generate multiple versions of a lambda function? i.e. as the following pseudo-code:
[] __stdcall (){};
[] __cdecl (){}; etc.
如果调用约定没有定义,如何在调用lambda函数之后正确回收堆栈空间?
Lambdas
[...]另外,在Visual Studio 2012中的Visual C ++中,无状态lambda可转换为函数指针。 [(Visual Studio 2012中的Visual C ++甚至比这更好),因为我们已经将无状态的lambda转换为具有任意调用约定的函数指针。当使用期望类似<$ )
c $ c> __ stdcall 函数指针。 编辑:
注意:调用转换不在C ++ Standard中,它取决于其他规范,如平台ABI(应用程序二进制接口)。
以下答案基于输出汇编代码,其中 / FAs编译器选项。
所以这只是一个猜测,请向微软询问更多细节; P
Q1。什么是C ++ lambda函数的调用约定?
Q3。如果调用约定没有定义,如何在调用lambda函数后正确回收堆栈空间?
首先,C ++ lambda(-expression)不是函数(也不是函数指针),你可以像调用普通函数一样调用 operator()
给lambda对象。
并且输出汇编代码表示VC ++ 2012通过 __ thiscall
调用转换生成lambda-body
。如何指定C ++ lambda函数的调用约定?
AFAIK,没有办法。 (可能只有 __ thiscall
)
Q4。编译器是否自动生成多个版本的lambda函数?即如下面的伪代码:[...]
可能是 这是一个例子; The following code was compiled with VC++ 2012: I think the errors are normal yet the OKs are abnormal. So, my questions are:
On VC++ 2012, compiler choose automatically calling conversion for stateless lambdas (that has no capture variables) when you convert "stateless lambda to function pointer". Lambdas [...] Additionally in Visual C++ in Visual Studio 2012, stateless lambdas are convertible to function pointers. [...] (The Visual C++ in Visual Studio 2012 is even better than that, because we've made stateless lambdas convertible to function pointers that have arbitrary calling conventions. This is important when you are using APIs that expect things like
EDITED: NB: The calling conversion is out of C++ Standard, it depends on other specification such as platform ABI(application binary interface). The following answers are based on output assembly code with /FAs compiler option.
So it's a mere guess, and please ask Microsoft for more detail ;P Q1. What's the calling convention of a C++ lambda function? Q3. If the calling convention is not defined, how to correctly recycle the stack space after having called a lambda function? First of all, C++ lambda(-expression) is NOT a function (nor function pointer), you can call Q2. How to specify the calling convention of a C++ lambda function? AFAIK, there is no way. (It may be only Q4. Does the compiler automatically generate multiple versions of a lambda function? i.e. as the following pseudo-code: [...] Probably No.
The VC++ 2012 lambda-type provides only one lambda-body implementation ( Here is an example;
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VC ++ 2012 lambda类型提供只有一个lambda-body实现( void operator()()
),但为每个调用转换提供了多个用户定义的转换为函数指针 void(__fastcall *)(void),
void(__stdcall *)(void)
和 void(__cdecl *)(void)
type)。
//输入源代码
auto lm = [](){/ * lambda-body * /};
//从VC ++ 2012反转C ++代码输出汇编代码
class lambda_UNIQUE_HASH {
void __thiscall operator()(){
/ * lambda-body * /
}
//用户定义的转换
typedef void(__fastcall * fp_fastcall_t)();
typedef void(__stdcall * fp_stdcall_t)();
typedef void(__cdecl * fp_cdecl_t)();
operator fp_fastcall_t(){...}
operator fp_stdcall_t(){...}
operator fp_cdecl_t(){...}
};
lambda_UNIQUE_HASH lm;
void f1(void (__stdcall *)())
{}
void f2(void (__cdecl *)())
{}
void __cdecl h1()
{}
void __stdcall h2()
{}
int main()
{
f1(h1); // error C2664
f2(h2); // error C2664
f1([](){}); // OK
f2([](){}); // OK
auto fn = [](){};
f1(fn); // OK
f2(fn); // OK
}
__stdcall
function pointers.)
operator()
to lambda object like a calling normal function.
And output assembly code says that VC++ 2012 generates lambda-body with __thiscall
calling conversion.
__thiscall
)
void operator()()
), but provides multiple "user-defined conversion to function pointer" for each calling conversion (operator return function pointer with void (__fastcall*)(void)
, void (__stdcall*)(void)
, and void (__cdecl*)(void)
type).// input source code
auto lm = [](){ /*lambda-body*/ };
// reversed C++ code from VC++2012 output assembly code
class lambda_UNIQUE_HASH {
void __thiscall operator()() {
/* lambda-body */
}
// user-defined conversions
typedef void (__fastcall * fp_fastcall_t)();
typedef void (__stdcall * fp_stdcall_t)();
typedef void (__cdecl * fp_cdecl_t)();
operator fp_fastcall_t() { ... }
operator fp_stdcall_t() { ... }
operator fp_cdecl_t() { ... }
};
lambda_UNIQUE_HASH lm;