Question

我正在尝试在Xcode中构建一个非常简单的命令行应用程序，它将打印出有关MXF视频文件的基本信息。为此，我需要在这里使用libmxf，libbmx和libbmx库：

http://sourceforge.net/p/bmxlib/home/Home/

此时我的C ++代码非常简单：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cerrno>
#include <vector>

#include <bmx/mxf_reader/MXFFileReader.h>
#include <bmx/mxf_reader/MXFGroupReader.h>
#include <bmx/mxf_reader/MXFSequenceReader.h>
#include <bmx/mxf_reader/MXFFrameMetadata.h>
#include <bmx/MXFUtils.h>
#include <bmx/Utils.h>

using namespace std;
using namespace bmx;

#define MXF_OPEN_READ(fn, pf)   mxf_disk_file_open_read(fn, pf)

int main(int argc, const char * argv[])
{
    std::vector<const char *> filenames;
    std::cout << "mxfheader: execution beginning...\n";
    for (int cmdln_index = 0; cmdln_index < argc; cmdln_index++) {
        if (!check_file_exists(argv[cmdln_index])) {
            if (argv[cmdln_index][0] == '-') {
                fprintf(stderr, "Unknown argument '%s'\n", argv[cmdln_index]);
            } else {
                fprintf(stderr, "Failed to open input filename '%s'\n", argv[cmdln_index]);
            }
            return 1;
        }
        filenames.push_back(argv[cmdln_index]);
    }

    std::cout << filenames[0] << "\n";
    return 0;
}

当我编译BMX库时，我确信使用64位支持运行configure，如下所示：

./configure --build=x86_64-apple-darwin11.4.2 --host=x86_64-apple-darwin11.4.2 CFLAGS="-arch x86_64" CXXFLAGS="-arch x86_64" LDFLAGS="-arch x86_64" CC=clang CXX=clang++

在Build Settings下的XCode Project中，我已将/ usr / local / lib添加到我的搜索路径中。在Build Phases下，我将“libbmx-0.1.3.dylib”，“libMXF-1.0.4.dylib”和“libMXF ++ - 1.0.4.dylib”添加到“Link Binary With Libraries”部分。

我已经确认这些库确实是64位（文件libbmx-0.1.3.dylib返回libbmx-0.1.3.dylib：Mach-O 64位动态链接共享库x86_64）。

每次我尝试构建应用程序时，都会收到以下链接器错误：

Ld /Users/ned/Library/Developer/Xcode/DerivedData/mxfheader-bkwawmplsoqpdadfxartceqkbolo/Build/Products/Debug/mxfheader normal x86_64
    cd /Users/ned/Documents/src/mxfheader
    setenv MACOSX_DEPLOYMENT_TARGET 10.7
    /Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/clang++ -arch x86_64 -isysroot /Applications/Xcode.app/Contents/Developer/Platforms/MacOSX.platform/Developer/SDKs/MacOSX10.8.sdk -L/Users/ned/Library/Developer/Xcode/DerivedData/mxfheader-bkwawmplsoqpdadfxartceqkbolo/Build/Products/Debug -L/usr/local/lib -F/Users/ned/Library/Developer/Xcode/DerivedData/mxfheader-bkwawmplsoqpdadfxartceqkbolo/Build/Products/Debug -filelist /Users/ned/Library/Developer/Xcode/DerivedData/mxfheader-bkwawmplsoqpdadfxartceqkbolo/Build/Intermediates/mxfheader.build/Debug/mxfheader.build/Objects-normal/x86_64/mxfheader.LinkFileList -mmacosx-version-min=10.7 -stdlib=libc++ -lbmx-0.1.3 -lMXF-1.0.4 -lMXF++-1.0.4 -o /Users/ned/Library/Developer/Xcode/DerivedData/mxfheader-bkwawmplsoqpdadfxartceqkbolo/Build/Products/Debug/mxfheader

Undefined symbols for architecture x86_64:
  "bmx::check_file_exists(std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> >)", referenced from:
      _main in main.o
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)

任何帮助将不胜感激。谢谢！

Answer 1

您的问题是命令行中的选项-stdlib=libc++。它导致链接到错误的libc ++，你需要将其设为-stdlib=libstdc++，因为这是编译libbmx库的stdlib。

在C ++标准库的Apple LLVM compiler选项下，选择：libstdc++，或选择compiler default（也应选择libstdc ++）

使用外部库构建Xcode项目时链接器错误

1 个答案: