【发布时间】:2019-04-01 14:29:30
【问题描述】:
将数据输入数组的问题。
我正在使用 Dev C++ 执行和编译。下面是我的代码问题所在:
for (int k=0;k<=149;k++){ // k - scanning through Y within single file #i
for (int l=0;l<=199;l++){ // l - scanning through X within single file #i
inp >> X[l]>> Y[k]>> Z>> Mx[l][k]>> My[l][k]>> Mz[l][k];
Z 没有缺少下标,它就是这样设计的。
完整的错误代码:
[Error] no match for 'operator>>' (operand types are
'std::basic_istream<char>::__istream_type {aka std::basic_istream<char>}' and 'double [2]')
欢迎任何建议谢谢。
完整代码:
#pragma warning(disable:4786)
#include <cstdio>
#include <iostream>
#include <fstream>
#include <cstdlib>
#include <string>
#include <cstring>
using namespace std;
string filename(int filenumber);
int main () {
char *flnmptr;
string str,tmpr;
char flnm[50];
!!! make sure that the array dimensions for input data are right - that might screw the output files
dimensions are written in the 2nd header line of the input file
double X[150], Y[200], Z[2], Mx[150][200][2], My[150][200][2], Mz[100][100][2], time; // data from file
// double hX[5001], vY[5001], hMx[5001],hMy[5001],hMz[5001], vMx[5001], vMy[5001], vMz[5001]; // data to be written to file
//int xcnt, ycnt; // variable to count how many values will be written to a file
const int N=600; //number of files
//const int ptN=20; // number of points in the line
const double timestep=1e-14; // time step of files
const double X0=2.749313e-1; // X coordinate of line to read into vertical stripe output
const double Y0=-9.614638e-1; // Y coordinate of line to read into horizontal stripe output
ofstream outp("SWstripeCoupledCorner50difference10_120GHz.txt"); //opening output file
if (!outp){
cout << "Can't open file.\n";
return 1;
}
cout<<"Reading file #: ";
outp.setf (ios::scientific ); // set the output format
for (int i=1;i<=N;i++){
std::strcpy(flnm,"");
tmpr=filename(i);
flnmptr=&tmpr[0];
std::strcpy(flnm,flnmptr);
ifstream inp;
inp.open(flnm); // opens the file
if(!inp) // if file couldn't be opened
{
cerr << "Error: file could not be opened" << endl;
exit(1);
}
std::getline(inp,str,'\n');
std::getline(inp,str,'\n');
inp.setf (ios::scientific);
for (int k=0;k<=149;k++){ // k - scanning through Y within single file #i
for (int l=0;l<=199;l++){ // l - scanning through X within single file #i
inp >> X[l]>> Y[k]>> Z>> Mx[l][k]>> My[l][k]>> Mz[l][k];
}
}
}
}
【问题讨论】:
-
请粘贴完整代码。
inp是什么类型? -
X、Y、Z、Mx、My、Mz的具体类型是什么? -
请提供minimal reproducible example。从错误看来,您正在尝试使用
operator>>读取数组。没有这样的内置运算符,您应该考虑在循环中读取此类数组的元素。 -
好的。我会发布完整的代码
-
啊,我想我明白了。我将深入研究解释数组和循环的文献,并尝试应用于我的代码。谢谢。