Question

我在Windows 7 Pro 64位上运行mingw-w64。

我在命名空间中尝试代码时遇到错误，但没有命名空间就可以正常工作。代码是：

//foo.h

#ifndef FOO_H
#define FOO_H

namespace joke {

    extern "C" int fooStack[10];
    extern "C" int *fooPtr;
    extern "C" void fooInit();
    extern "C" void fooPush(int iVar);
    extern "C" int fooPop();

}

#endif // FOO_H

// foo.cpp

#include "foo.h"

using namespace joke;

int fooStack[10];
int *fooPtr;

void fooInit() {
    fooPtr = &fooStack[0] + 9;
}

void fooPush(int iVar) {
    *fooPtr = iVar;
    fooPtr -= 1;
}

int fooPop() {
    int oVar;
    fooPtr += 1;
    oVar = *fooPtr;
}

// bar.cpp

#include <iostream>
#include "foo.h"

using namespace std;
using namespace joke;

int inVar = 0;
int outVar = 0;

void Report() {
    int i;
    cout << "Pointer:   " << fooPtr << endl;
    cout << "Stack:     ";
    for (i = 0; i < 10; i++) {
        cout << fooStack[i] << " ";
    }
    cout << endl;
    cout << "inVar:     " << inVar << endl;
    cout << "outVar:    " << outVar << endl << endl;
}

int main() {
    int i;
    fooInit();
    cout << endl;
    cout << "Initializing" << endl;
    Report();
    for (i = 0; i < 8; i++) {
        inVar = (i * 34) + 34;
        cout << "set inVar" << endl;
        Report();
        fooPush(inVar);
        cout << "Push inVar" << endl;
        Report();
    }
    for (i = 0; i < 8; i++) {
        outVar = fooPop();
        cout << "Pop outVar" << endl;
        Report();
    }
    return 0;
}

makefile：

all:

    g++ -S foo.cpp

    g++ -S bar.cpp

    g++ -c foo.s

    g++ -c bar.s

    g++ -o FB foo.o bar.o

正如我所说，如果删除或注释掉命名空间规范，这一切都可以正常工作。使用命名空间规范后，结果如下：

c:\work\gccWork\FooTest>make
g++ -S foo.cpp
foo.cpp: In function 'void fooInit()':
foo.cpp:11:2: error: reference to 'fooPtr' is ambiguous
  fooPtr = &fooStack[0] + 9;
  ^
foo.cpp:8:6: note: candidates are: int* fooPtr
  int *fooPtr;
       ^
In file included from foo.cpp:3:0:
foo.h:9:18: note:                 int* joke::fooPtr
  extern "C" int *fooPtr;
                  ^
foo.cpp:11:12: error: reference to 'fooStack' is ambiguous
  fooPtr = &fooStack[0] + 9;
            ^
foo.cpp:7:5: note: candidates are: int fooStack [10]
 int fooStack[10];
     ^
In file included from foo.cpp:3:0:
foo.h:8:17: note:                 int joke::fooStack [10]
  extern "C" int fooStack[10];
                 ^
foo.cpp: In function 'void fooPush(int)':
foo.cpp:15:3: error: reference to 'fooPtr' is ambiguous
  *fooPtr = iVar;
   ^
foo.cpp:8:6: note: candidates are: int* fooPtr
 int *fooPtr;
      ^
In file included from foo.cpp:3:0:
foo.h:9:18: note:                 int* joke::fooPtr
  extern "C" int *fooPtr;
                  ^
foo.cpp:16:2: error: reference to 'fooPtr' is ambiguous
  fooPtr -= 1;
  ^
foo.cpp:8:6: note: candidates are: int* fooPtr
 int *fooPtr;
      ^
In file included from foo.cpp:3:0:
foo.h:9:18: note:                 int* joke::fooPtr
  extern "C" int *fooPtr;
                  ^
foo.cpp: In function 'int fooPop()':
foo.cpp:21:2: error: reference to 'fooPtr' is ambiguous
  fooPtr += 1;
  ^
foo.cpp:8:6: note: candidates are: int* fooPtr
 int *fooPtr;
      ^
In file included from foo.cpp:3:0:
foo.h:9:18: note:                 int* joke::fooPtr
  extern "C" int *fooPtr;
                  ^
foo.cpp:22:10: error: reference to 'fooPtr' is ambiguous
  oVar = *fooPtr;
          ^
foo.cpp:8:6: note: candidates are: int* fooPtr
 int *fooPtr;
      ^
In file included from foo.cpp:3:0:
foo.h:9:18: note:                 int* joke::fooPtr
  extern "C" int *fooPtr;
                  ^
make: *** [all] Error 1

我已经在线查看了各种c ++教科书和参考资料，以及一些参考资料和论坛。我没有看到语法有任何问题。

我错过了什么？

Answer 1

您要求C链接。但是C没有命名空间概念。这对我来说似乎是一种利益冲突。

Answer 2

全局（在未命名的fooPtr）和namespace中声明了

namespace joke。它们是不同的实体。

using namespace joke然后告诉编译器使用joke中的名称作为匹配标识符的候选者

当代码具有语句::fooPtr时，编译器可以看到名称joke::fooPtr和fooPtr = &fooStack[0] + 9。 using namespace joke表示::fooPtr和joke::fooPtr都是有效匹配。没有理由优先选择其中一个，因此编译器会因模糊而拒绝代码。

C ++名称空间错误

2 个答案: