How to append a char to a String in C using a function-CodePudding

I was writing a lexical analyzer in which I need to append a char to a string (a char *). For some reason, the code below is resulting in string having a value of "(null)" when I print it to stdout. The function is given below.

void append_char(char *buffer, char c) {
  if(buffer == NULL) {
    buffer = malloc(sizeof(char));
    if(buffer == NULL) {
      fprintf(stderr, "COuld not allcocate memory to buffer\n");
    }
  } else {
    buffer = realloc(buffer, sizeof(buffer)   sizeof(char));
  } 
  buffer[sizeof(buffer) - 1] = c;
}

When I run the lines

 char *buf = NULL;
 append_char(buf, 'a');
 append_char(buf, '\0');
 printf("buffer: %s\n", buf);

it prints (null) to stdout. How can I fix this?

CodePudding user response：

Pass by value

append_char(char *buffer, char c) does not affect the caller's buf in main(): append_char(buf, 'a');. buf remains NULL. This leads to OP's output.

Insufficient size

Insufficient size for the newly allocated string. No room for the null character.

Wrong size

With char *buffer, sizeof(buffer) is the size of a pointer, not the amount allocated beforehand.

Lost memoery

When buffer = realloc(buffer, sizeof(buffer) sizeof(char)); fails (realloc() returns NULL) , the original value of buffer is lost. Save the result and test.

Note: OK to call realloc(NULL, ...).

char *append_char(char *buffer, char c) {
  size_t old_length = buffer ? strlen(buffer) : 0;
  size_t new_length = old_length   1; //  1 for c
  // Size needed for a string is its length   1
  char *new_buffer = realloc(buffer, new_length   1); //  1 for \0
  if (new_buffer == NULL) {
    fprintf(stderr, "Could not allocate memory to buffer\n");
    free(buffer);
    return NULL;
  }
  new_buffer[old_length] = c; 
  new_buffer[old_length   1] = '\0'; 
  return new_buffer;
}

// Usage
buf = append_char(buf, 'a');