文件输入在C + + [英] File Input in C++

问题描述

我试着在互联网上搜索一段时间，但是我可以用C ++找到文件输入。编写一个程序来执行文件中添加2个大于0的数字，但不使用scanf或cin。我想把文件加载到内存中，但是我能找到的所有代码都需要知道文件名。该文件在一行中用2个整数格式化，由一个空格分隔，并且有多行整数。程序将输出这两个数字的总和。我可以很容易地使用scanf做到这一点，但如果我得到一个巨大的文件，我想将它加载到内存（稍后保存映射）。

加载文件进入内存是给我的麻烦，因为我不知道文件名，也不知道如何查找，除非用户输入文件的名称（不会发生）。我希望程序像这样被执行，但使用最原始的，基本的C ++ IO形式：

  ./ myprog < boatloadofnumbers 
  

我如何开始我的程序，以整个boatloadofnumbers作为文件，所以我可以使用更多的基本功能，如read（）？ （还有，上面的方法是什么？传入输入？）

  int main（）{
 int a， b; 
 while（scanf（％i，％i，& a，& b）> -1）{
 printf（％i \ n，（a + b）） ; 
} // endwhile 
 return 0; 
} // endmain 
  

解决方案

清楚在不知道文件名的情况下如何阅读文件。大概你不知道编译时间的文件名。没关系，你可以从命令行在 runtime 中得到这个，就像这样：

 。 / myprog boatloadofnumbers 
  

然后你的文件名在 argv [1] ，你可以使用 std :: ifstream 来访问它。

通过直接在 stdin 上输入重定向 （例如 ./ myprog< boatloadofnumbers ），您根本不需要文件名， code> std :: cin 。

以下 main（）将处理这两种情况：

$ pre $ int
if（argc == 2）
{
std :: cerr<< 从文件中读取<< argv [1]<<的std :: ENDL;
std :: ifstream ifs（argv [1]）;
if（ifs）
{
sum_lines（ifs）;
}
else
{
std :: cerr<< 无法读取<< argv [1]<<的std :: ENDL;
}
}
else
{
std :: cerr<< 从标准输入读取<<的std :: ENDL;
sum_lines（std :: cin）;




$ sum_lines（） 可能看起来有点像这样：
$ b $ $ $ $ p $ void $ sum_lines（std :: istream& is）
{
int first = 0，second = 0;
std :: string line =;
while（std :: getline（is，line））
{
std :: istringstream iss（line）;
if（>>> first>> second）
{
std :: cout<<第一<< +<<第二< =<<第一+第二<<的std :: ENDL;
}
else
{
std :: cerr<< 无法解析[<<行<< ]<<的std :: ENDL;



code
$ b

这并不关心在那里输入，所以你可以很容易地注入一个 std :: istringstream 来进行单元测试。此外，这不会将整个文件读入内存，每次只读一行，所以它应该处理 averybigboatloadofnumbers 。

I've been searching the internet for a while, but all I can find for file input in C++ is when you know the filename.

I'm trying to write a program to perform an addition of 2 numbers that are greater than 0 from a file, but without using scanf or cin. I want to load the file into memory, but all of the code I can find in regards to this situation requires knowledge of the filename. The file is formatted with 2 integers on a single line, separated by a space, and there are multiple lines of integers. The program will output the sum of the two numbers. I can easily do this with scanf, but if I were given a massive file, I would want to load it into memory (save mapping for later).

Loading the file into memory is giving me trouble, because I do not know the filename, nor how to find out, unless the user inputs the name of the file (not going to happen). I want the program to be executed like so, but using the most raw, and basic forms of C++ IO:

./myprog < boatloadofnumbers

How would I start my program to take the whole "boatloadofnumbers" as a file, so I can use more basic functions like read()? (also, what is the above method called? passing input?)

int main(){
    int a,b;
    while (scanf("%i,%i",&a,&b)>-1){
        printf("%i\n",(a+b));
    } //endwhile
    return 0;
} //endmain

解决方案

It's not entirely clear how you're going to read a file when you don't know the filename. Presumably you don't know the filename at compile-time. That's okay, you can get this from the command-line at runtime, like this:

./myprog boatloadofnumbers

Then your filename is in argv[1] and you can access it using a std::ifstream.

If you're being given the input directly on stdin via redirection (such as ./myprog < boatloadofnumbers) you don't need a filename at all, you can just use std::cin.

The following main() will deal with both of these situations:

int main(int argc, char* argv[])
{
    if (argc == 2)
    {
        std::cerr << "Reading from file" << argv[1] << std::endl;
        std::ifstream ifs(argv[1]);
        if (ifs)
        {
            sum_lines(ifs);
        }
        else
        {
            std::cerr << "Could not read from " << argv[1] << std::endl;
        }
    }
    else
    {
        std::cerr << "Reading from stdin" << std::endl;
        sum_lines(std::cin);
    }
}

A sample sum_lines() may look a bit like this:

void sum_lines(std::istream& is)
{ 
    int first = 0, second = 0;
    std::string line = "";
    while (std::getline(is, line))
    {
        std::istringstream iss(line);
        if (is >> first >> second)
        {
            std::cout << first << " + " << second << " = " << first + second << std::endl;
        }
        else
        {
            std::cerr << "Could not parse [" << line << "]" << std::endl;
        }
    }
}

This doesn't care from where the input comes, so you can easily inject a std::istringstream for unit-testing. Also, this doesn't read the whole file into memory, just one line at a time, so it should deal with averybigboatloadofnumbers.

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