如何避免多重定义链接错误?

发布于 2024-10-11 12:20:54 字数 2012 浏览 9 评论 0原文

除了将 hello() 函数移动到另一个源 (.cpp) 文件或重命名该函数之外。还有其他方法可以避免链接错误吗?

staticLibA.h

#ifndef _STATIC_LIBA_HEADER
#define _STATIC_LIBA_HEADER

int hello(void);
int hello_staticLibA_only(void);

#endif

staticLibA.cpp

#include "staticLibA.h"

int hello(void)
{
    printf("\nI'm in staticLibA\n");
    return 0;
}

int hello_staticLibA_only(void)
{
    printf("\nstaticLibA: hello_staticLibA_only\n");
    return 0;
}

输出:

g++ -c -Wall -fPIC -m32 -o staticLibA.o staticLibA.cpp
ar -cvq ../libstaticLibA.a staticLibA.o
a - staticLibA.o

staticLibB.h

#ifndef _STATIC_LIBB_HEADER
#define _STATIC_LIBB_HEADER

int hello(void);
int hello_staticLibB_only(void);

#endif

staticLibB.cpp

#include "staticLibB.h"

int hello(void)
{
    printf("\nI'm in staticLibB\n");
    return 0;
}

int hello_staticLibB_only(void)
{
    printf("\nstaticLibB: hello_staticLibB_only\n");
    return 0;
}

<强>输出:

g++ -c -Wall -fPIC -m32 -o staticLibB.o staticLibB.cpp 
ar -cvq ../libstaticLibB.a staticLibB.o 
a - staticLibB.o

ma​​in.cpp

extern int hello(void);
extern int hello_staticLibA_only(void);
extern int hello_staticLibB_only(void);

int main(void)
{
  hello();
  hello_staticLibA_only();
  hello_staticLibB_only();
  return 0;
}

输出:

g++ -c  -o main.o main.cpp
g++ -o multipleLibsTest main.o  -L. -lstaticLibA -lstaticLibB -lstaticLibC -ldl -lpthread -lrt
./libstaticLibB.a(staticLibB.o): In function `hello()':
staticLibB.cpp:(.text+0x0): multiple definition of `hello()'
./libstaticLibA.a(staticLibA.o):staticLibA.cpp:(.text+0x0): first defined here
collect2: ld returned 1 exit status
make: *** [multipleLibsTest] Error 1

Beside moving the hello() function into another source (.cpp) file or renaming the function. Is there any other methods to avoid the linking error?

staticLibA.h

#ifndef _STATIC_LIBA_HEADER
#define _STATIC_LIBA_HEADER

int hello(void);
int hello_staticLibA_only(void);

#endif

staticLibA.cpp

#include "staticLibA.h"

int hello(void)
{
    printf("\nI'm in staticLibA\n");
    return 0;
}

int hello_staticLibA_only(void)
{
    printf("\nstaticLibA: hello_staticLibA_only\n");
    return 0;
}

output:

g++ -c -Wall -fPIC -m32 -o staticLibA.o staticLibA.cpp
ar -cvq ../libstaticLibA.a staticLibA.o
a - staticLibA.o

staticLibB.h

#ifndef _STATIC_LIBB_HEADER
#define _STATIC_LIBB_HEADER

int hello(void);
int hello_staticLibB_only(void);

#endif

staticLibB.cpp

#include "staticLibB.h"

int hello(void)
{
    printf("\nI'm in staticLibB\n");
    return 0;
}

int hello_staticLibB_only(void)
{
    printf("\nstaticLibB: hello_staticLibB_only\n");
    return 0;
}

output:

g++ -c -Wall -fPIC -m32 -o staticLibB.o staticLibB.cpp 
ar -cvq ../libstaticLibB.a staticLibB.o 
a - staticLibB.o

main.cpp

extern int hello(void);
extern int hello_staticLibA_only(void);
extern int hello_staticLibB_only(void);

int main(void)
{
  hello();
  hello_staticLibA_only();
  hello_staticLibB_only();
  return 0;
}

output:

g++ -c  -o main.o main.cpp
g++ -o multipleLibsTest main.o  -L. -lstaticLibA -lstaticLibB -lstaticLibC -ldl -lpthread -lrt
./libstaticLibB.a(staticLibB.o): In function `hello()':
staticLibB.cpp:(.text+0x0): multiple definition of `hello()'
./libstaticLibA.a(staticLibA.o):staticLibA.cpp:(.text+0x0): first defined here
collect2: ld returned 1 exit status
make: *** [multipleLibsTest] Error 1

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评论(2

谁人与我共长歌 2024-10-18 12:20:54

由于您似乎拥有这两个库,因此不清楚为什么您无法重命名该函数...

在您的 main 中,您有这一行:

hello();

如果您进行链接,您期望在这里发生什么错误消失了吗?它应该调用 LibALibB 中的实现吗?依靠将库传递给链接器的顺序来确定链接哪个函数似乎是一个非常糟糕的主意。在真实的示例中,如果您的 hello_staticLibB_only 函数调用 hello() 会发生什么?它最终可能会调用其他库中的函数版本...

当您使用 g++ 时,您应该考虑将库函数放入 命名空间 (它们旨在帮助您避免这种命名冲突)。这将使您的代码和链接器能够区分方法之间的差异。

按照 LibA 的这种方法,您将获得:

staticLibA.h

#ifndef _STATIC_LIBA_HEADER
#define _STATIC_LIBA_HEADER

// Declare namespace to keep library functions together
namespace LibA {
    int hello(void);
    int hello_staticLibA_only(void);
}

#endif

staticLibA.cpp

#include "staticLibA.h"
#include <stdio.h>

// Indicate that contained function definitions belong in the LibA namespace
namespace LibA {
    int hello(void)
    {
        printf("\nI'm in staticLibA\n");
        return 0;
    }

    int hello_staticLibA_only(void)
    {
        printf("\nstaticLibA: hello_staticLibA_only\n");
        return 0;
    }
}

ma​​in.cpp

// These declarations would usually be in a header... but I've left
// them here to match your sample code...

// declare relevant functions to belong to the LibA namespace
namespace LibA{
    extern int hello(void);
    extern int hello_staticLibA_only(void);
}

// declare relevant functions from LibB (note they are not
// in a namespace)
extern int hello(void);
extern int hello_staticLibB_only(void);

int main(void)
{
    // Explicitly call the hello from LibA
    LibA::hello();
    // Call the other library function (also from LibA)
    LibA::hello_staticLibA_only();

    // Call library functions from LibB (note they don't require a namespace
    // because I haven't updated it to have one)
    hello();
    hello_staticLibB_only();
    return 0;
}

Since you appear to own both libraries, it's unclear why you can't rename the function...

In your main, you have this line:

hello();

What are you expecting to happen here, if you make the linking error go away? Should it call the implementation in LibA, or LibB? Relying on the order that you pass the libraries to the linker to determine which function gets linked seems like a very bad idea. In a real example, what would happen if your hello_staticLibB_only function was calling hello()? It could end up calling the version of the function that's in the other library...

As you're using g++, you should consider putting your library functions into a namespace (they're designed to help you avoid this kind of nameing conflict). This would allow both your code and the linker to tell the difference between the methods.

Following this approach for LibA, you would have:

staticLibA.h

#ifndef _STATIC_LIBA_HEADER
#define _STATIC_LIBA_HEADER

// Declare namespace to keep library functions together
namespace LibA {
    int hello(void);
    int hello_staticLibA_only(void);
}

#endif

staticLibA.cpp

#include "staticLibA.h"
#include <stdio.h>

// Indicate that contained function definitions belong in the LibA namespace
namespace LibA {
    int hello(void)
    {
        printf("\nI'm in staticLibA\n");
        return 0;
    }

    int hello_staticLibA_only(void)
    {
        printf("\nstaticLibA: hello_staticLibA_only\n");
        return 0;
    }
}

main.cpp

// These declarations would usually be in a header... but I've left
// them here to match your sample code...

// declare relevant functions to belong to the LibA namespace
namespace LibA{
    extern int hello(void);
    extern int hello_staticLibA_only(void);
}

// declare relevant functions from LibB (note they are not
// in a namespace)
extern int hello(void);
extern int hello_staticLibB_only(void);

int main(void)
{
    // Explicitly call the hello from LibA
    LibA::hello();
    // Call the other library function (also from LibA)
    LibA::hello_staticLibA_only();

    // Call library functions from LibB (note they don't require a namespace
    // because I haven't updated it to have one)
    hello();
    hello_staticLibB_only();
    return 0;
}
小傻瓜 2024-10-18 12:20:54

链接错误特指 hello.这是因为两个库都提供了“hello”的定义。这里没有其他链接错误。

您可以将 hello 放在单独的库中,让 hello 驻留在单独的库中,或者仅具有针对 hello 对象文件的可执行链接 [hello.o]

The linking error specifically refers to hello. This shows up because both libraries provide definitions of "hello". There is no other linking error here.

You can either put hello in a separate library, have hello reside in a separate library, or just have the executable link against a hello object file [hello.o]

~没有更多了~
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