[英]Read everything up to a whitespace from a text file
(C++) 我創建了一個函數來打開文本文件並將內容分配給一個數組。 數組中的前 2 個元素是網格的大小。 但是,如果前兩個數字中的一個或兩個都是兩位數,則不會將它們讀為兩位數。 有沒有辦法做到這一點?
int openMap()
{
std::string fileName;
std::cout << "Please enter the file name with extension that you want to open: ";
std::cin >> fileName;
system("CLS");
std::ifstream file(fileName); //OPENS MAP FILE
int tmp;
int i = 0;
if (!file.is_open()) //CHECKS IF THE MAP FILE HAS OPENED CORRECTLY
{
std::cout << "Error Occured!\nCould not open file.";
return 0;
}
while (!file.eof()) //READS THE MAP FILE AND PASSES THE INFORMATION INTO AN ARRAY
{
file >> tmp;
checkNumber(tmp);
if (valid == true) //IF THE CHARACTER IS NOT A NUMBER THEN IT WONT BE PASSED INTO THE ARRAY
{
tmpArray[i] = tmp;
i++;
valid = false;
}
row = tmpArray[1]; //ASSIGNS THE FIRST 2 NUMBERS OF THE MAP FILE TO ROW AND COL VARIABLES
col = tmpArray[0];
}
return row, col;
}
我認為我必須重寫
file >> tmp
以某種不同的方式,但不確定如何。 有沒有辦法掃描文本文件直到它遇到空格?
文本文件內容如下所示
6 4 0 0 1 0 0 0 2 0 1 0 1 1 0 0 1 0 0 0 0 0 0 0 3 0
(6 或 4 或兩者都可以是兩位數)
編輯:
for (int j = 0; j < row; j++)
{
for (int k = 0; k < col; k++)
{
_map[j][k] = tmpArray[l];
std::cout << _map[j][k] << " ";
l++;
}
}
代碼中有相當多的錯誤,您可能應該使用調試器逐步檢查並確定程序的哪些部分未按預期運行。
while(!file.eof())
file >> tmp;
checkNumber(tmp);
if (valid == true) //IF THE CHARACTER IS NOT A NUMBER THEN IT WONT BE PASSED INTO THE ARRAY
{
tmpArray[i] = tmp;
i++;
valid = false;
}
row = tmpArray[1]; //ASSIGNS THE FIRST 2 NUMBERS OF THE MAP FILE TO ROW AND COL VARIABLES
col = tmpArray[0];
您在循環的每次迭代中都設置row=tmpArray[1]
和col = tmpArray[0]
,這不僅是不必要的,而且是不正確的,特別是因為 row=tmpArray[1] 是在i=0
時執行的,而沒有放入任何內容tmpArray[1]
呢。
編輯:由於較少的變量和類型轉換,這更小,更不容易出錯,並且更易於閱讀:
int row,col;
//Add error checking here
cin >> col;
cin >> row;
cout << "Cols: " << col << " Rows: " << row << endl;
vector<vector<int> >_map(row, vector<int>(col,0));
for(int j=0; j < row; j++)
{
for(int k=0; k < col; k++)
{
int tmp;
cin >> tmp;
//Add error checking for tmp
_map[j][k] = tmp;
cout << _map[j][k] << endl;
}
}
您的代碼存在一些問題。 首先,您的函數的返回類型是int
但您正在返回多個值。 這是一個完整的運行代碼,可以解決您的問題。
#include <iostream>
#include <fstream>
#include <vector>
std::vector< std::vector<int> > openMap() {
std::string fileName;
std::cout << "Please enter the file name with extension that you want to open: ";
std::cin >> fileName;
std::fstream myfile(fileName, std::ios_base::in);
int row, col;
myfile >> row;
myfile >> col;
int a;
std::vector< std::vector<int> > retval;
for (int i = 0; i < row; i++) {
std::vector<int> v1;
for (int j = 0; j < col; j++) {
myfile >> a;
v1.push_back(a);
}
retval.push_back(v1);
}
return retval;
}
int main(int argc, char * argv[])
{
std::vector< std::vector<int> > _map = openMap();
for(int i = 0; i < _map.size(); i++) {
for (int j = 0; j < _map[i].size(); j++) {
std::cout << _map[i][j] << " ";
}
std::cout << std::endl;
}
return 0;
}
估計沒多少人會感興趣。 但請參閱下面的可能解決您的問題的方法。
該代碼使用現代 C++ 算法。
它非常簡單明了。
#include <iostream>
#include <fstream>
#include <vector>
#include <iterator>
#include <algorithm>
int main() {
// Ask user, to give a filename
std::cout << "Please enter the file name with extension that you want to open: ";
// Get the filename from the user
if (std::string fileName; std::cin >> fileName) {
// Open the file and check, if it is open
if (std::ifstream sourceFile(fileName); sourceFile) {
// Read the number of rows and columns of the matrix
if (size_t numberOfColumns, numberOfRows; sourceFile >> numberOfColumns >> numberOfRows) {
// Create a matrix with the given number of rows and columns
std::vector<std::vector<int>> result(numberOfRows, std::vector<int>(numberOfColumns, 0));
// Read data from the input stream and put it into the matrix
for (size_t i = 0; i < numberOfRows; ++i) {
std::copy_n(std::istream_iterator<int>(sourceFile), numberOfColumns, result[i].begin());
}
// Print result. Go through all lines and then copy line elements to std::cout
std::for_each(result.begin(), result.end(), [](std::vector<int>& c) {
std::copy(c.begin(), c.end(), std::ostream_iterator<int>(std::cout, " ")); std::cout << "\n"; });
}
}
else {
std::cerr << "\n*** Error: Could not open source File\n\n";
}
}
return 0;
}
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