将int转换为char * C

Question

So I am trying to read the words from a file. However, I have to use putchar(ch) where ch is an int . How do I convert ch to a string (char *) so I can store it in a char * variable and pass it to another function that takes char * as a parameter. And I actually just want to store it but not print it.

This is what I have:

int main (void)
{
   static const char filename[] = "file.txt";
   FILE *file = fopen(filename, "r");
   if ( file != NULL )
   {
      int ch, word = 0;
      while ( (ch = fgetc(file)) != EOF )
      {
         if ( isspace(ch) || ispunct(ch) )
         {
            if ( word )
            {
               word = 0;
               putchar('\n');
            }
         }
         else
         {
            word = 1;
            putchar(ch);
         }
      }
      fclose(file);
   }
   return 0;
}

Answer 1

sprintf(char_arr, "%d", an_integer);

This makes char_arr equal to string representation of an_integer (This doesn't print anything to console output in case you're wondering, this just 'stores' it) An example:

char char_arr [100];
int num = 42;
sprintf(char_arr, "%d", num);

char_arr now is the string "42" . sprintf automatically adds the null character \\0 to char_arr .

If you want to append more on to the end of char_arr, you can do this:

sprintf(char_arr+strlen(char_arr), "%d", another_num);

the '+ strlen' part is so it starts appending at the end.

more info here: http://www.cplusplus.com/reference/cstdio/sprintf/

Answer 2

So you have a single value of char type, aka int8_t (or uint8_t on some systems). You have it stored in an int , so fgetc can return -1 for error, but still be able to return any 8bit character.

Single characters are just 8-bit integers, which you can store in any size of integer variable without problems. Put them in an array with a zero-byte at the end, and you have a string.

char buffer[10] = {0};
int c = 'H';
buffer[0] = c;
// now buffer holds the null-terminated string "H"
buffer[1] = 'e';
buffer[2] = 'l';  // you can see where this is going.
c = buffer[1];  // c = 'e' = 101
  // (assuming you compile this on a system that uses ASCII / unicode, not EBCDIC or some other dead character mapping).

Note that the string-terminating zero-bytes got into buffer because I initialized it. Using an array initializer zeroes any elements you don't mention in your initializer list.

Answer 3

To represent a single character as a character string, I find using a simple 2-character buffer to be as easy as anything else. You can take advantage of the fact that dereferencing the string points to the first character and simply assign the character you wish to represent as a string. If you have initialized your 2-char buffer to 0 (or '\\0' ) when declared, you have insured your string is always null-terminated :

Short Example

#include <stdio.h>

int main (void) {

    int ch;
    char s[2] = {0};
    FILE *file = stdin;

    while ( (ch = fgetc(file)) != EOF ) {
        *s = ch;
        printf ("ch as char*: %s\n", s);
    }

    return 0;
}

Use/Output

$ printf "hello\n" | ./bin/i2s2
ch as char*: h
ch as char*: e
ch as char*: l
ch as char*: l
ch as char*: o
ch as char*:

Note: you can add && ch != '\\n' to the while condition to prevent printing the newline.