我又遇到了问题。我能够正确编译我的DLL。但是一旦我开始将它链接到一个可以编译的可执行文件。发生了一些事情。
这是CMD:
>gcc -c -o dllmain.o dllmain.c -D ADD_EXPORTS
>gcc -o dllmain.dll dllmain.o -s -shared -Wl,--subsystem,windows
这部分结果很好,产生了一个看似好看的dllmain.dll。
以下是问题的开始:
>gcc -c -o dllmain.o dllmain.c
>gcc -o main.exe -s main.o -L. -ladd
c:/mingw/bin/../lib/gcc/mingw32/4.8.1/../../../../mingw32/bin/ld.exe: cannot find -ladd
collect2.exe: error: ld returned 1 exit status
有人可以告诉我它出现的原因吗? (所有文件名都经过三重检查,并且是正确的!)
这是“dllmain.c”
/* Replace "dll.h" with the name of your header */
#include "dll.h"
#include <windows.h>
#include <stdio.h>
DLLIMPORT void HelloWorld()
{
MessageBox(0,"Hello World from DLL!\n","Hi",MB_ICONINFORMATION);
}
DLLIMPORT CALL add(int a, int b)
{
return a + b;
}
DLLIMPORT CALL subtract(int a, int b)
{
return a - b;
}
DLLIMPORT CALL multiply(int a, int b)
{
return a * b;
}
DLLIMPORT CALL divide(int a, int b)
{
if(b = 0)
{
MessageBox(0, "Cannot divide by zero!", "Division Error", 1);
return;
}
return a / b;
}
DLLIMPORT CALL modulo(int a, int b)
{
if(b = 0)
{
MessageBox(0, "Cannot divide by zero!", "Modulo Error", 1);
}
}
BOOL WINAPI DllMain(HINSTANCE hinstDLL,DWORD fdwReason,LPVOID lpvReserved)
{
switch(fdwReason)
{
case DLL_PROCESS_ATTACH:
{
break;
}
case DLL_PROCESS_DETACH:
{
break;
}
case DLL_THREAD_ATTACH:
{
break;
}
case DLL_THREAD_DETACH:
{
break;
}
}
/* Return TRUE on success, FALSE on failure */
return TRUE;
}
这是“dll.h”
#if ADD_EXPORTS
#define DLLIMPORT __declspec(dllexport)
#else
#define DLLIMPORT __declspec(dllimport)
#endif
#define CALL __cdecl
#ifdef __cplusplus
extern "C"
{
#endif
DLLIMPORT void HelloWorld();
DLLIMPORT CALL add(int a, int b);
DLLIMPORT CALL subtract(int a, int b);
DLLIMPORT CALL multiply(int a, int b);
DLLIMPORT CALL divide(int a, int b);
DLLIMPORT CALL modulo(int a, int b);
#ifdef __cplusplus
}
#endif
最后,“main.c”
#include <stdio.h>
#include <stdlib.h>
#include "dll.h"
main()
{
int a, b, c;
char chc;
printf("Input 2 numbers: ");
printf("Input operation: ");
scanf("%c", &chc);
switch(chc)
{
case '+': addition(a, b); break;
case '-': subtraction(a, b); break;
case '*': multiply(a, b); break;
case '/': divide(a, b); break;
case '%': modulo(a, b); break;
}
getch();
}
答案 0 :(得分:1)
只需使用:
gcc -o main.exe main.c dllmain.dll
链接器查找libadd.a
,在您的情况下不存在
同样,对于编译,您不必提供dllmain.o