无法将int打印到文件中-Can'tprintaninttoafile

3  

The definition of write is as follows:

写的定义如下:

ssize_t write(int fd, const void *buf, size_t count);

The definition of snprintf is as follows:

snprintf的定义如下:

int snprintf(char *str, size_t size, const char *format, ...);

Now you're feeding snprintf to where write is expecting void *buf. i.e. the address to where the data is.

现在你正在将snprintf提供给write期望void * buf的地方。即数据所在的地址。

So you'll have to do

所以你必须这样做

snprintf(buffer2, 10, "%d", listBoard[2]);

then

write(file_write, buffer2, 18);

EDIT:

That will fix the syntax error. Now semantically you can see that you're writing 10 bytes to buffer2 but you're writing 18 to the file. This will fill the file with junk. So i suggest.

这将修复语法错误。现在在语义上你可以看到你正在向buffer2写入10个字节,但是你正在写18个文件。这将用垃圾填充文件。所以我建议。

int len = snprintf(buffer2, 10, "%d", listBoard[2]);
write(file_write, buffer2, len);

EDIT 2:

int len = snprintf(buffer2, 1, "%d \n", ' ');

First off your format specifier has 3 bytes while you're only writing 1 byte to the buffer2. Second, why is your argument ' ' when you're supposed to give the listBoard[n]?

首先,您的格式说明符有3个字节,而您只向缓冲区2写入1个字节。第二,为什么你的论点''你应该给listBoard [n]?

2  

Use dprintf() to write formatted strings to files:

使用dprintf()将格式化的字符串写入文件:

dprintf(file_write, "%d", listBoard[2]);

dprintf will handle all the write stuff for you, so this is the only call you need with open and close.

dprintf将为您处理所有写入内容,因此这是您打开和关闭时唯一需要的调用。

dprintf is the fprintf for file descriptors. More informations here: https://linux.die.net/man/3/dprintf

dprintf是文件描述符的fprintf。更多信息:https://linux.die.net/man/3/dprintf

0  

The snprintf (and family) functions returns an integer.

snprintf(和family)函数返回一个整数。

You use this integer in the write call as a pointer to a buffer. That will lead to undefined behavior.

您在write调用中使用此整数作为指向缓冲区的指针。这将导致不确定的行为。

Call snprintf separately, and then pass buffer2 as the second argument. And use strlen to get the actual length of the string in buffer2, not hard-code the magic number 18 which will lead to reading from out-of-bounds and another case of UB.

分别调用snprintf,然后传递buffer2作为第二个参数。并使用strlen来获取buffer2中字符串的实际长度,而不是硬编码将导致从越界读取和另一个UB情况的幻数18。

0  

First of all snprintf returns an integer that is the number of chars that would have been written if the string were not truncated (as you can read on the snprintf manual)

首先,snprintf返回一个整数,该整数是字符串未被截断时已写入的字符数(正如您可以在snprintf手册中看到的那样)

you cannot then use it in the write command. You can then use:

你不能在write命令中使用它。然后你可以使用:

 if ( snprintf(buffer2, 10,"%d",listBoard[2]) >= 10 ) {
    /* handle truncation error*/
 }
 if(write(file_write,buffer2,18)!=18){
     err_sys("write_error");

Anyway since buffer2 is maximum 10 characters why do you expect to write 18 bytes?

无论如何,因为buffer2最多10个字符,你为什么要写18个字节?

Other possible error: buffer2 should be at least of 10 + 1 byte for null termination.

其他可能的错误:对于空终止,buffer2应至少为10 + 1字节。

0  

You never specify that you must use raw file descriptors, so let's port your code up one level to C's standard library's FILE abstraction. It's easier to work with.

您从未指定必须使用原始文件描述符,因此让我们将您的代码上移一级到C的标准库的FILE抽象。它更容易使用。

void printToFile(const int listBoard[]) {
  FILE * const out = fopen("board.txt", "wt");
  if (out == NULL) {
    err_sys("error output file");
  }
  for(int j = 0; j <3; ++j) {
    for (int i = 0; i <3; ++i)
      fprintf(out, "%d ", listBoard[3 * j + i]);
    fprintf(out, "\n");
  }
  fclose(out);
}

Note that this uses loops to iterate over the elements of the board array, this seemed to be missing from your original code and part of the problem you had with it.

请注意,这使用循环来迭代板数组的元素,这似乎是原始代码中缺少的,也是您使用它的部分问题。

This prints:

1 2 3 
4 5 6 
7 8 9 

There's a trailing space on each line due to me being lazy, but hopefully that's fine. Also I omitted some inner error checking.

由于我很懒,每条线都有一个尾随空间,但希望这很好。我也省略了一些内部错误检查。