c++ - 处理标准一维数组时忽略前导空格

Question

在这种情况下，我们必须从一个名为 small.txt 的文件中读取。该文件的内容如下所示:

NOT      11010011
AND      10010010  11001110
OR       10011001  11100101
CONVERT  10010110
LSHIFT   11001101  3
WRONG    01010100  10101010

在这个文件中，每个单词都代表一个新行的开始。我遇到的问题是让我的代码在最后(第 3 列)中读取。每列的长度为十个空格。以下是我必须遵守的规则:

根据文件中的命令，对命令后面的操作数进行运算。每个命令都相互独立，因此一个命令不会影响另一个。

这些命令都是二进制命令，但是你不能为此使用 C++ 的内置命令。这意味着没有:“~、&、|、<<(用作二元运算符时)”。你当然可以使用&&和||，它们是不同的。 << 可以像往常一样用于 cout，只是不能用作二元运算符。

您不能使用字符串来读入操作数。

我试过:

inputFile.ignore(20, '\n')

但如果这是正确的方法。我没有正确使用它。

这是我的代码:

/*
       ====================================================================
                                                                                     SUMMARY 

Read commands which will require the program to perform some operation on either one or two bit 
patterns, determine the result of the operation, and output accordingly. 

NOT: 

Takes 1 operand and performs a bitwise logical NOT. At each position, if the operand has a 0, 
the result will contain a

1. If the operand has a 1, the result will contain a 0.

Eg. operand 11010011
result 00101100

In logical operations, a 1 represents TRUE and a 0 FALSE.

AND:

Takes 2 operands and performs a bitwise logical AND. At each position, if both operand 1 and 
operand 2 contain a 1, the result will be a 1. Otherwise the result is 0.

Eg. operand 1 10010010
operand 2 11001110
result 10000010

OR:

Takes 2 operands and performs a bitwise logical OR. At each position, if either operand 1 or 
2 or both contain a 1, the result will contain a 1. Otherwise the result will contain a 0 
(inclusive OR).

Eg. operand 1 10011001
operand 2 11100101
result 11111101

CONVERT:

Takes 1 operand and converts it to a base 10 integer. Note: we will let every bit in these 
binary numbers represent part of a positive binary integer, i.e. there is no "sign" bit. 
Thus we can only represent positive integers in the range from 0 thru (28 - 1).

Eg. operand 10010110
result = (1 * 2**7) + (0 * 2**6) + (0 * 2**5) + (1 * 2**4)
+ (0 * 2**3) + (1 * 2**2) + (1 * 2**1) + (0 * 2**0) = 150 in base 10

LSHIFT:

Logical Shift to Left Takes 1 operand and an integer N as input. The bit values are shifted N 
positions to the left. Data "pushed off" the left end is lost. Zeroes replace the lost bits.

Eg. operand 1 11001101 N = 3
result 01101000
You may assume N is valid, i.e. 0 <= N <= 8

   ======================================================================
                                                                                   ASSUMPTIONS 

The binary operands will contain exactly 8 bits, where a bit is a binary digit. A byte contains 
8 bits.

Check for invalid command names. Assume that the binary operands are all correctly given in the 
data file.

   ======================================================================
                                                                                      INPUT 

From the data file binaryData.txt.

   =======================================================================
                                                                                     OUTPUT 

Echo print all input values. Then output in a suitable fashion the results of the operation
performed, and any necessary error messages.

*/

/* ========================================================================================*/
/*                                           HEADER FILES                              */

#include <iomanip>                                 // needed for output manipulation
#include <iostream>                    // needed for standard I/O routines
#include <fstream>
#include <string>                                 // needed for reading data from files
using namespace std;

/* ====================================================================================*/
/*                   FUNCTION                                                         /* ================================================================== */
/*            NAMED GLOBAL CONSTANTS                                                   */             

    const int ARRAY_SIZE = 8;                        // array size  

/* =========================================================================== */
/*       MAIN FUNCTION                                                          */
    int main (){

int numbers[ARRAY_SIZE];                        // array with 8 elements
int secondArray[ARRAY_SIZE];                    // array two with 8 elements
int thirdArray[ARRAY_SIZE];                    // array three with 8 elements 
int count = 0;                                 // loop counter variable

ifstream inputFile;                           // input file stream object

// open the file
inputFile.open ("small.txt");

// exit if a fatal error occurs opening the file
if( !inputFile ){

    cout << "Error: Data file could not be opened \n";
    system ("pause");
    return (EXIT_FAILURE);

} // end of not in file if statement

// read the array

string word;

while(inputFile){

    // stores the word read in by the file 
    inputFile >> word;

    if(word == "NOT"){

   cout << word << "     ";

   // This allows you to be able to read in each number one by one
   for(int i = 0; i < ARRAY_SIZE; i++){ 

       char letter(20);

       inputFile >> letter;
       letter = letter - '0';
       numbers[i] = static_cast<int>(letter);

       if (letter == 0){

           numbers[i] = letter + 1;
       }
       else{

           numbers[i] = letter - 1;
       }

    } // end of first for loop 

    for(int i = 0; i < ARRAY_SIZE; i++){

       cout << numbers[i];
     } // end of second for loop 
    } // end of if word == not check

    if(word == "AND"){

   cout << "\n\n" << word << "     ";

   // This allows you to be able to read in each number one by one
   for(int i = 0; i < ARRAY_SIZE; i++){ 

       //inputFile.ignore(20, '\n');

       char letter;

       inputFile.ignore(10) >> letter;
       letter = letter - '0';
       secondArray[i] = static_cast<int>(letter);

    } // end of first for loop 

    for(int i = 0; i < ARRAY_SIZE; i++){

       cout << secondArray[i];
     } // end of second for loop 
    } // end of if word == not check
} // end of while inputFile loop

inputFile.clear ( );
inputFile.close ( );



cout << endl;
system ("pause");
return (0);

    } // end of main function

Answer 1

您的代码已经使用了流中的前两列，并且流的位置超过了这些列。反过来，不必调用 ignore 来将流位置移动到这些列之后。除了忽略ifstream提供了一些您可能会觉得有用的附加功能。