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我在以下一段 C++ 代码中有一个错误,我无法弄清楚。任何人都可以帮忙。我有下面的代码及其输出。最好的。

double* alloc_Array=new double[m_int_NumChann*m_int_NumSamples];
int int_SizeArray=int(0);
std::ifstream File;
File.open(m_char_Address);
if(File.is_open()){
    std::cout<<"input file opened...\n";
    int i=int(0);
    do{
        File>>alloc_Array[i];
        i++;
    }while(!File.eof());
    int_SizeArray=i;
}else{
    std::cerr<<"ERROR: input file can't be opened.\n";
    system("pause");
}
File.close();
if((m_int_NumChann*m_int_NumSamples)!=int_SizeArray){
    std::cerr<<"WARNING: number of samples multiplied by number of channels is not equal to total data points in the input file:\n";
    std::cerr<<"       number of samples in each channel = "<<m_int_NumSamples<<'\n';
    std::cerr<<"       number of channels = "<<m_int_NumChann<<'\n';
    std::cerr<<"       total data points by multiplication = "<<m_int_NumSamples*m_int_NumChann<<'\n';
    std::cerr<<"       number of data points in the input file = "<<int_SizeArray<<'\n';
    system("pause");
}

输出:

   input file opened...

   WARNING: number of samples multiplied by number of channels is not equal to tota
   l data points in the input file:

   number of samples in each channel = 77824

   number of channels = 11

   total data points by multiplication = 856064

   number of data points in the input file = 856065

   Press any key to continue . . .
4

2 回答 2

3

解决此问题的最简单方法是不循环播放eof().

eof()尝试循环或正确循环存在众所周知的问题good(),请阅读以下问题示例:为什么循环条件内的 iostream::eof 被认为是错误的?测试 stream.good() 或 !stream.eof() 读取最后一行两次

您可以重新排序代码,使其仅i在成功读取值时递增:

int i=int(0);
while (File >> alloc_Array[i]) {
    i++;
}
int_SizeArray=i;
于 2014-07-22T16:10:57.357 回答
2

在您的 do{} while() 循环中,您每次都在递增 i 。考虑零长度文件的情况。将发生循环的第一遍,之后 i 将等于 1。由于将立即到达 EOF,因此不会发生后面的遍。然而,在这种情况下实际上没有发现任何样本。

您将希望在循环结束后将 i 递减一次。请记住计算找到的样本数量(比循环运行的次数少一)和尝试填充的数组中的元素数量之间的区别。

于 2014-07-22T15:17:43.897 回答