c - 从 C 中的命令行参数打开文件

Question

我希望我的 C 程序要求用户键入他们想要打开的文件的名称，并将该文件的内容打印到屏幕上。到目前为止，我正在学习 C 教程并拥有以下代码。但是当我执行它时，它实际上不允许我输入文件名。 (我得到“按任意按钮继续”，我正在使用代码块)

我在这里做错了什么？

#include <stdio.h>

int main ( int argc, char *argv[] )
{
    printf("Enter the file name: \n");
    //scanf
    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 );
        }
    }
    return 0;
}

Answer 1

这将从标准输入读取文件名。如果文件名未作为命令行的一部分提供，您可能只想执行此操作。

int main ( int argc, char *argv[] ) 
{ 
    char filename[100];
    printf("Enter the file name: \n"); 
    scanf("%s", filename);

    ...
    FILE *file = fopen( filename, "r" );