我有一个汇编例程，它以通用的方式调用一个已知使用stdcall约定并返回float的函数。此函数正被编组框架用于向脚本语言公开stdcall函数。

背景

以下是使用GNU内联汇编的函数，它在MinGW 4.3、Win32上编译：

inline uint64_t stdcall_invoke_return_float(int args_size_bytes,
                                            const char * args_ptr,
                                            void * func_ptr)
{
    uint64_t result;
    assert(
        0 == args_size_bytes % 4
            || !"argument size must be a multiple of 4 bytes");
#if defined(__GNUC__)
    asm
    (
        /* INPUT PARAMS:  %0 is the address where top of FP stack to be stored
         *                %1 is the number of BYTES to push onto the stack, */
        /*                   and during the copy loop it is the address of */
        /*                   the next word to push */
        /*                %2 is the base address of the array */
        /*                %3 is the address of the function to call */
            "testl %1, %1    # If zero argument bytes given, skip nt"
            "je    2f        # right to the function call.        nt"
            "addl  %2, %1n"
        "1:nt"
            "subl  $4, %1    # Push arguments onto the stack in   nt"
            "pushl (%1)      # reverse order. Keep looping while  nt"
            "cmp   %2, %1    # addr to push (%1) > base addr (%2) nt"
            "jg    1b        # Callee cleans up b/c __stdcall.    n"
        "2:nt"
            "call  * %3      # Callee will leave result in ST0    nt"
            "fsts  %0        # Copy 32-bit float from ST0->result"
        : "=m" (result)
        : "r" (args_size_bytes), "r" (args_ptr), "mr" (func_ptr)
        : "%eax", "%edx", "%ecx" /* eax, ecx, edx are caller-save */, "cc"
    );
#else
#pragma error "Replacement for inline assembler required"
#endif
    return result;
}

这只是一点胶水，使编写测试用例更容易：

template<typename FuncPtr, typename ArgType>
float float_invoke(FuncPtr f, int nargs, ArgType * args)
{
    uint64_t result = stdcall_invoke_return_float(
        nargs * sizeof(ArgType),
        reinterpret_cast<const char *>(args),
        reinterpret_cast<void *>(f)
    );
    return *reinterpret_cast<float *>(&result);
}

现在我有一些调用这个函数的测试用例：

__stdcall float TestReturn1_0Float()
{ return 1.0f; }
__stdcall float TestFloat(float a)
{ return a; }
__stdcall float TestSum2Floats(float a, float b)
{ return a + b; }
static const float args[2] = { 10.0f, -1.0f };
assert_equals(1.0f, float_invoke(TestReturn1_0Float, 0, args)); // test 1
assert_equals(10.0f, float_invoke(TestFloat, 1, args));         // test 2
assert_equals(-1.0f, float_invoke(TestFloat, 1, args + 1));     // test 3
assert_equals(9.0f, float_invoke(TestSumTwoFloats, 2, args));   // test 4

问题

随机地，测试3给我垃圾输出，而不是返回-1.0。

我想知道我是不是

• 未能在call指令之前保留某些状态
• 用fsts指令扰乱某些状态
• 从根本上误解了如何从返回float的stdcall函数中获得float值

非常感谢所有的帮助。