使用'/ n'从文本文件读取

Question

Okay. So I'm reading and storing text from a text file into a char array, this is working as intended. However, the textfile contains numerous newline escape sequences. The problem then is that when I print out the string array with the stored text, it ignores these newline sequences and simply prints them out as "\\n".

Here is my code:

char *strings[100];

void readAndStore(FILE *file) {
  int count = 0;
  char buffer[250];

  while(!feof(file)) {
    char *readLine = fgets(buffer, sizeof(buffer), file);
    if(readLine) {
      strings[count] = malloc(sizeof(buffer));
      strcpy(strings[count], buffer);
      ++count;
    }

  }
}

int main() {

  FILE *file1 = fopen("txts", "r");
  readAndStore(&*file1);
  printf("%s\n", strings[0]);
  printf("%s\n", strings[1]);
  return 0;
}

And the output becomes something like this:

Lots of text here \\n More text that should be on a new line, but isn't \\n And so \\n on and and on \\n

Is there any way to make it read the "\\n" as actual newline escape sequences or do I just need to remove them from my text file and figure out some other way to space out my text?

Answer 1

No. Fact is that \\n is a special escape sequence for your compiler, which turns it into a single character literal, namely "LF" (line feed, return), having ASCII code 0x0A . So, it's the compiler which gives a special meaning to that sequence.

Instead, when reading from file, \\n is read as two distinct character, ASCII codes 0x5c,0x6e .

You will need to write a routine which replaces all occurences of \\\\n (the string composed by characters \\ and n, the double escape is necessary to tell the compiler not to interpret it as an escape sequence) with \\n (the single escape sequence, meaning new line).

Answer 2

If you only intend to replace '\\n' by the actual character, use a custom replacement function like

void replacenewlines(char * str)
{
   while(*str)
   {
      if (*str == '\\' && *(str+1) == 'n') //found \n in the string. Warning, \\n will be replaced also.
      {
         *str = '\n'; //this is one character to replace two characters
         memmove(str, str+1, strlen(str)); //So we need to move the rest of the string leftwards
               //Note memmove instead of memcpy/strcpy. Note that '\0' will be moved as well
      }
      ++str;
   }
}

This code is not tested, but the general idea must be clear. It is not the only way to replace the string, you may use your own or find some other solution.

If you intend to replace all special characters, it might be better to lookup some existing implementation or sanitize the string and pass it as the format parameter to printf . As the very minimum you will need to duplicate all '%' signs in the string.

Do not pass the string as the first argument of printf as is, that would cause all kinds of funny stuff.