[英]Reading data from a text file with different variable types in C/C++
[英]Loading different data types from text file
我是StackExchange和C ++的新手,因此,如果我對問題的描述不夠好,則表示歉意。 我需要一些作業幫助。 我正在嘗試找到一種加載此文件並存儲其數據的方法。 該文件具有不同的數據類型,因此到目前為止,我想到的是使用結構數組存儲所有值,並使用三重嵌套的for循環將數據保存在正確的變量中。
.dat文件的外觀如下,帶有“注釋”以幫助描述正在發生的事情。
100A 2 // model number, 2 different versions of that model
0 0 0 // number of quarters, dimes, and nickels
5 //number of items in the vending machine
1A 1034 5 // Code combonation, ID Number, Quantity
1B 1000 10
1C 1100 10
1D 1123 20
1E 1222 5
0 0 0 // number of quarter, dimes, nickels in 2nd model
7 // number of items in the second version of that model
1A 2180 20
1B 1283 20
1C 3629 5
1D 3649 3
1E 4051 15
1F 4211 1
1G 5318 5
100B 3 // New model, with 3 different versions of itself.
2 10 5 //everything repeats like model 100A
7
這是我想出的代碼
#include <fstream>
#include <iostream>
#include <iomanip>
#include <string>
using namespace std;
struct VMdata
{
ifstream inFile;
string model[1];
int version[1];
int q[5];
int d[5];
int n[5];
int size[5];
string id[30];
int code[30];
int num[30];
char dummy;
};
int main()
{
VMdata New;
cout << fixed << setprecision(2) << showpoint;
New.inFile.open("machines.dat");
cout << "Model Data"
<< endl << endl;
int count1 = 0;
int count2= 0;
int count3 = 0;
for (int i = 0; i < 2; i ++)
{
cout << "i :" << count1 << endl;
New.inFile >> New.model[i] >> New.version[i]; // loads model number and number of versions i times
count1 = count1 + 1;
for (int j = 0; j < New.version[count1 -1]; j ++)
{
cout <<"j :" << count2 << endl;
New.inFile >> New.q[count2] >> New.d[count2] >> New.n[count2] >> New.size[count2]; // loads number of q, d, n, j times
count2 = count2 + 1;
for (int k = 0; k < New.size[count2 - 1]; k++)
{
cout << "k :" << count3 << endl;
New.inFile >> New.id[count3] >> New.code[count3] >> New.num[count3]; // loads id number, code number, and total number k times
count3 = count3 + 1;
}
}
}
New.inFile.close();
count1 = 0;
count2= 0;
count3 = 0;
cout << endl;
for ( int i = 0; i < 2; i ++)
{
cout << New.model[i] << setw(12) << New.version[i] << endl << endl;
count1 = count1 + 1;
for (int j = 0; j < New.version[count1 -1]; j ++)
{
cout << New.q[count2] << setw(12) << New.d[count2] << setw(12) << New.n[count2] << endl << setw(12) << New.size[count2] << endl;
count2 = count2 + 1;
for (int k = 0; k < New.size[count2 - 1]; k++)
{
cout << New.id[count3] << setw(12) << New.code[count3] << setw(12) << New.num[count3] << endl << endl;
count3 = count3 + 1;
}
}
}
return 0;
}
這是我的測試輸出。
100A 2
3 15 0 // should be 0, 0, 0
5
1A 1034 5
1B 1000 10
1C 1100 10
1D 1123 20
1E 1222 5
0 0 0
7
1A 2180 20
1B 1283 20
1C 3629 5
1D 3649 3
1E 4051 15
1F 4211 1
1G 5318 5
♥ ☻ 3 // the heart and smile should be 100B lol
2 10 5
7
1A 2180 10
1B 1283 10
1C 3629 5
1D 3649 3
1E 4051 15
1F 4211 10
1G 3026 5
5 6 3
6
1A 6626 5
1B 6155 5
1C 5982 10
1D 5573 3
1E 5454 10
1F 5336 50
10 10 10
5
1A 1034 5
1B 1000 5
1C 1100 5
1D 1123 5
1E 1210 12
Press any key to continue . . .
如您所見,對於第一台機器,四分之一,一角硬幣和鎳幣的數量是錯誤的,而對於第二台機器,型號名稱是錯誤的。 如果有人有任何建議,將不勝感激。
看來您正在破壞數據。 您的結構僅分配一個模型和一個版本,但i索引的值為0和1。第二次讀取這些值時,該地址將覆蓋您的硬幣計數。
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