문제
이것은 내가 묻는 또 다른 질문과 관련이 있지만, 그것은 전적으로 자신의 질문이지만.
컴파일하면 두 가지 오류가 발생합니다.
1>. asst4.cpp (73) : Error C2065 : 'Outfile': 노출되지 않은 식별자
1>. asst4.cpp (73) : 오류 C2228 : 왼쪽의 '.Close'는 클래스/struct/Union이 있어야합니다.
내가 여기서 잘못한 일에 대해 약간 혼란스러워? 권장이나 아이디어가 있습니까? (실제 의상은 코드의 맨 위에 있습니다.
전체 코드는 다음과 같습니다.
#include<iostream>
#include<fstream> //used for reading/writing to files.
#include<string> //needed for the filename.
#include<stdio.h> //for goto statement
using namespace std;
int main()
{
string start;
char choice;
char letter;
int x;
int y;
int z;
string filename;
int garbage = rand()%('!' - '~' + 1 );
cout << "Would you like to encrypt or decrypt a file? Please type enc, dec, or stop (case sensitive): " ;
cin >> start;
while(start == "enc")
{
x = 1;
y = 1;
cout << "How many garbage characters would you like between each correct character?: " ;
cin >> z;
cout << endl << "Please insert the name of the document you wish to encrypt, make sure you enter the name, and the file type (ie: filename.txt): " ;
cin >> filename;
ifstream infile(filename.c_str());
while(!infile.eof())
{
ofstream outfile("encrypted.txt", ios::out);
infile.get(letter);
if (x == y)
{
outfile << garbage;
x++;
}
else
{
if((x - y) == z)
{
outfile << letter;
y = x;
}
else
{
outfile << garbage;
x++;
}
}
}
cout << endl << "Encryption complete...please return to directory of program, a new file named encrypted.txt will be there." << endl;
infile.close();
outfile.close();
cout << "Do you wish to try again? Please press y then enter if yes (case sensitive).";
cin >> choice;
if(choice == 'y')
{
start = "enc";
}
else
{
cout << endl << "Do you wish to decrypt a file? Please press y then enter if yes (case sensitive).";
if(choice = 'y')
{
start == "dec";
}
else
{
start == "no";
}
}
}
while(start == "dec")
{
//lets user choose whether to do another document or not.
//used to track each character in the document.
x = 1; //first counter for tracking correct letter.
y = 1; //second counter (1 is used instead of 0 for ease of reading, 1 being the "first character").
//third counter (used as a check to see if the first two counters are equal).
//allows for user to input the filename they wish to use.
cout << "Please make sure the document is in the same file as the program, thank you!" << endl << "Please input document name: " ;
cin >> filename; //getline(cin, filename);
cout << endl;
cout << "'Every nth character is good', what number is n?: ";
cin >> z; //user inputs the number at which the character is good. IE: every 5th character is good, they would input 5.
cout << endl;
z = z - 1; //by subtracting 1, you now have the number of characters you will be skipping, the one after those is the letter you want.
ifstream infile(filename.c_str()); //gets the filename provided, see below for incorrect input.
if(infile.is_open()) //checks to see if the file is opened.
{
while(!infile.eof()) //continues looping until the end of the file.
{
infile.get(letter); //gets the letters in the order that that they are in the file.
if (x == y) //checks to see if the counters match...
{
x++; //...if they do, adds 1 to the x counter.
}
else
{
if((x - y) == z) //for every nth character that is good, x - y = nth - 1.
{
cout << letter; //...if they don't, that means that character is one you want, so it prints that character.
y = x; //sets both counters equal to restart the process of counting.
}
else //only used when more than every other letter is garbage, continues adding 1 to the first
{ //counter until the first and second counters are equal.
x++;
}
}
}
cout << endl << "Decryption complete...please return to directory of program, a new file named encrypted.txt will be there." << endl;
infile.close();
cout << "Do you wish to try again? Please press y then enter if yes (case sensitive).";
cin >> choice;
if(choice == 'y')
{
start == "dec";
}
else
{
cout << endl << "Do you wish to encrypt a file? Please press y then enter if yes (case sensitive).";
if(choice == 'y')
{
start == "enc";
}
else
{
start == "no";
}
}
}
else //this prints out and program is skipped in case an incorrect file name is used.
{
cout << "Unable to open file, please make sure the filename is correct and that you typed in the extension" << endl;
cout << "IE:" << " filename.txt" << endl;
cout << "You input: " << filename << endl;
cout << "Do you wish to try again? Please press y then enter if yes (case senstive)." ;
cin >> choice;
if(choice == 'y')
{
start == "dec";
}
else
{
start == "no";
}
}
getchar(); //because I use visual C++ express.
}
}
미리 감사드립니다! 제프
해결책
범위 문제. 당신은 당신의 while 루프 내에서 Outfile을 선언하고 있지만, Said While Loop 외부에서 액세스하려고합니다.
이동하다 ofstream outfile("encrypted.txt", ios::out);
당신의 바로 다음 줄로 ifstream infile(filename.c_str());
당신 앞에 있습니다 while(!infile.eof())
.
다른 팁
X -istence에서 언급 한 것처럼 두 가지 선택이 있습니다. Outfile.close ()에 대한 호출이 컴파일되지 않은 이유는 'Outfile'이 Outfile 선언과 동일한 범위에 존재하지 않기 때문입니다.
당신은 줄을 움직일 수 있습니다
ofstream outfile("encrypted.txt", ios::out);
while 루프 외부에서 infile과 동일한 범위 규칙을 순종하거나 leclople.close ()로 호출을 이동할 수 있습니다. while 루프 내부는이를 전파가 존재하는 현재 범위로 이동합니다.
파일을 여는 것은 상당히 비싼 작업이기 때문에 Outfile 선언을 while 루프 외부에서 옮기는 것이 선호하는 것이 좋습니다. 한 번 열면이 경우에 한 번 닫으십시오.