Question

这是一个简单的程序，应该复制一个内容文件名为copyme的文件here。我通过以下命令创建了copyme，其中包含一些文本：

touch copyme.txt
open copyme.txt

然后我键入文本，并保存文件 touch copyme.txt命令。

然后我编写了一个程序：

// Program to copy one file ot another

#include <stdio.h>

int main (void)
{
    char in_name[64], out_name[64];
    FILE *in, *out;
    int c;

    // get file names from user

    printf("Enter name of file to be copied: ");
    scanf("%63s", in_name);

    printf("Entere name of output file: ");
    scanf("%63s", out_name);

    // open input and output files

    if ( (in = fopen(in_name, "r")) == NULL)
    {
        printf("Can't open %s for reading.\n", in_name);
        return 1;
    }

    if ( (out = fopen(out_name, "w")) == NULL)
    {
        printf("Can't open %s for writing.\n", out_name);
        return 2;
    }

    while ((c = getc(in)) != EOF)
        putc(c, out);

    // Close open files

    fclose (in);
    fclose (out);

    printf("File has been copied\n");

    return 0;
}

然后在终端跑了。这是输出：

Enter name of file to be copied: copyme
Entere name of output file: here
Can't open copyme for reading.

编译器不识别copyme文件，尽管它是物理存在于文件夹中（我看到了，我打开它，我读了它）。我很感激你的帮助。我是这个新手。谢谢！

Answer 1

变化

   if ( (in = fopen(in_name, "r")) == NULL)
    {
        printf("Can't open %s for reading.\n", in_name);
        return 1;
    }

到

   #include <errno.h>
   if ( (in = fopen(in_name, "r")) == NULL)
    {

        perror("Can't open file for reading.\n");
        return 1;
    }

您将收到一条人类可读的消息，告诉您为何无法读取该文件

我的副本文件功能没有按预期工作

1 个答案: