为什么visual spion 2015 sp3不能生成此行的代码

时间:2017-04-12 05:06:18

标签: visual-c++-2015

我注意到Visual Studio社区2015 sp3奇怪地没有为​​单行创建代码(在代码中显示)(调试和发布)。我最小化了项目并将代码放在这里。

#include "stdafx.h"

enum AMINZIPCMD { AMINZIP_RELEASE = 0 };
enum AMINZIPRESULT { AMINZIP_SUCCESS = 0, AMINZIP_FAILURE };
struct AMINZIP {
    unsigned __int32 m_cbSize;
    unsigned __int32 m_dwType;
};
#define DLL_EXPORT __declspec(dllexport) __stdcall
extern "C" {
    int DLL_EXPORT uAminZip(enum AMINZIPCMD iCmd, struct AMINZIP * pAminZip);
};

#define global_func

/*------------------------------------------------------------*/
global_func int DLL_EXPORT uAminZip(enum AMINZIPCMD iCmd, struct AMINZIP * pAminZip)
{
    //No code generated for this line
    if (pAminZip->m_cbSize != sizeof(*pAminZip)) AMINZIP_FAILURE;
    if (pAminZip->m_dwType != 718624670) return AMINZIP_FAILURE;
    return AMINZIP_SUCCESS;
}

int APIENTRY wWinMain(HINSTANCE hInstance, HINSTANCE hPrevInstance, LPWSTR lpCmdLine, int nCmdShow)
{
    struct AMINZIP aminzip;
    return uAminZip(AMINZIP_RELEASE, &aminzip);
}


/*------------------------------------------------------------*/

和uAminZip函数的汇编窗口

global_func int DLL_EXPORT uAminZip(enum AMINZIPCMD iCmd, struct AMINZIP * pAminZip)
{
00F01690 55                   push        ebp  
00F01691 8B EC                mov         ebp,esp  
00F01693 81 EC C0 00 00 00    sub         esp,0C0h  
00F01699 53                   push        ebx  
00F0169A 56                   push        esi  
00F0169B 57                   push        edi  
00F0169C 8D BD 40 FF FF FF    lea         edi,[ebp-0C0h]  
00F016A2 B9 30 00 00 00       mov         ecx,30h  
00F016A7 B8 CC CC CC CC       mov         eax,0CCCCCCCCh  
00F016AC F3 AB                rep stos    dword ptr es:[edi]  
    if (pAminZip->m_cbSize != sizeof(*pAminZip)) AMINZIP_FAILURE;
    if (pAminZip->m_dwType != 718624670) return AMINZIP_FAILURE;
00F016AE 8B 45 0C             mov         eax,dword ptr [pAminZip]  
00F016B1 81 78 04 9E 57 D5 2A cmp         dword ptr [eax+4],2AD5579Eh  
00F016B8 74 07                je          uAminZip+31h (0F016C1h)  
00F016BA B8 01 00 00 00       mov         eax,1  
00F016BF EB 02                jmp         uAminZip+33h (0F016C3h)  
    return AMINZIP_SUCCESS;
00F016C1 33 C0                xor         eax,eax  
}
00F016C3 5F                   pop         edi  
00F016C4 5E                   pop         esi  
00F016C5 5B                   pop         ebx  
00F016C6 8B E5                mov         esp,ebp  
00F016C8 5D                   pop         ebp  
00F016C9 C2 08 00             ret         8  

我创建了一个新的Win32项目并将代码复制到它,但问题仍然存在

mr.abzadeh

1 个答案:

答案 0 :(得分:0)

该行是否包含“return”语句?