我準備並行化一些C++ OpenCV代碼。我遇到了段錯誤,我無法理解發生了什麼。爲什麼這個OpenCV代碼段錯誤?
下面是代碼:
class ExhaustiveComparisonMT
{
vector<Mat> *m_sigs;
Mat *m_dist;
public:
ExhaustiveComparisonMT(vector<Mat> sigs, Mat dist)
{
m_sigs = &sigs;
m_dist = &dist; // gdb breakpoint 1 here
}
void operator() (size_t last_row, size_t last_col) const
{
Mat diff = (*m_sigs)[0].clone(); // segfault happens here, gdb breakpoint 2 here
for (size_t i = 0; i != last_row; ++i)
for (size_t j = 0; j != last_row; ++j)
{
cv::absdiff((*m_sigs)[i], (*m_sigs)[j], diff);
m_dist->at<double>(i, j) = cv::sum(diff).val[0];
}
}
};
void
exhaustive_comparison(vector<Mat> sigs, Mat dist)
{
size_t width = sigs.size();
ExhaustiveComparisonMT ecmt(sigs, dist);
ecmt(width, width);
}
基本上,矩陣的矢量被傳遞給構造。指向矢量的指針保存爲成員變量,因此可以在exhaustive_comparison
中再次訪問該矢量。但是,該函數試圖訪問向量的第一個元素。
我試着用兩個斷點(見代碼)來診斷gdb的問題。在斷點1上:
(gdb) p (*m_sigs)[0]
$1 = (cv::Mat &) @0x7fffee77d010: {flags = 1124024325, dims = 2, rows = 1, cols = 712, data = 0x624ec0 "", refcount = 0x0, datastart = 0x624ec0 "", dataend = 0x6259e0 "",
datalimit = 0x6259e0 "", allocator = 0x0, size = {p = 0x7fffee77d018}, step = {p = 0x7fffee77d060, buf = {2848, 4}}}
因此,正確訪問第一個元素。現在,我們進入斷點2並嘗試相同的操作:
(gdb) p (*m_sigs)[0]
$2 = (cv::Mat &) @0x7fffee77d010: <error reading variable>
第一個元素似乎不再可訪問!它的地址是相同的(0x7fffee77d010)。這裏發生了什麼?
最後,如果我向前一步,我得到:
Program received signal SIGSEGV, Segmentation fault.
0x00007ffff7a0dd50 in cv::Mat::copyTo(cv::_OutputArray const&) const() from /usr/local/lib/libopencv_core.so.2.4
OpenCV的嘗試訪問的第一個元素克隆它,失敗。
爲什麼第一個元素在構造函數中可訪問,但在exhaustive_comparison
成員函數中不可訪問?
啊,我忘記了一切都通過C++的價值傳遞,除非明確指定。謝謝。 – misha 2013-04-09 06:55:26