如何在 C++ 中创建一个在命令行上给出名称的文件？

Question

我正在做一些 C++ 教程，到目前为止我非常擅长。然而，有一件事情让我的知识获取困惑、脱轨，让我很头疼。

如何创建一个在命令行中指定名称的文件？

Answer 1

您问的是如何从命令行获取字符串来命名要打开的文件？

#include <iostream>
#include <cstdlib>
#include <fstream>

int main(int argc,char *argv[]) {
    if(2>argc) {
        std::cout << "you must enter a filename to write to\n";
        return EXIT_FAILURE;
    }
    std::ofstream fout(argv[1]); // open a file for output
    if(!fout) {
        std::cout << "error opening file \"" << argv[1] << "\"\n";
        return EXIT_FAILURE;
    }
    fout << "Hello, World!\n";
    if(!fout.good()) {
        std::cout << "error writing to the file\n";
        return EXIT_FAILURE;
    }
    return EXIT_SUCCESS;
}

Answer 2

您需要解析命令行参数并使用其中之一作为文件的文件名。请参阅此代码：

#include <stdio.h>

int main ( int argc, char *argv[] )
{
    if ( argc != 2 ) /* argc should be 2 for correct execution */
    {
        /* We print argv[0] assuming it is the program name */
        printf( "usage: %s filename", argv[0] );
    }
    else 
    {
        // We assume argv[1] is a filename to open
        FILE *file = fopen( argv[1], "r" );

        /* fopen returns 0, the NULL pointer, on failure */
        if ( file == 0 )
        {
            printf( "Could not open file\n" );
        }
        else 
        {
            int x;
            /* read one character at a time from file, stopping at EOF, which
               indicates the end of the file.  Note that the idiom of "assign
               to a variable, check the value" used below works because
               the assignment statement evaluates to the value assigned. */
            while  ( ( x = fgetc( file ) ) != EOF )
            {
                printf( "%c", x );
            }
            fclose( file );
        }
    }
}

请参阅此处了解更多详细信息： http://www.cprogramming.com/tutorial /c/lesson14.html