Runtime Error in Dynamic Sized Character Array

Question

#include <stdio.h>
#include <stdlib.h>

int main(void) {
  int i = 0;
  char c, *input;
  input = (char *) malloc(sizeof(char));

  if(input == NULL) {
    printf("NOT ENOUGH SPACE!");
    exit(1);
  }

  printf("Input a string, press ENTER when done: ");

  while((c = getchar()) != '\n') {
    realloc(input, (sizeof(char)));
    input[i++] = c;
  }

  input[i] = '\0';
  printf("\nYou've entered the string: %s\n", input);
}

The above code snippet works well for small input. But it fails whenever the input provided is of large size. Either there is a run time error or a segmentation error. There may be some error in reallocation of memory space. I basically want to store a character array dynamically from user, i.e without mentioning the capacity of the input the user can directly put in any size of character array.

Paul R · Accepted Answer · 2017-04-11 21:12:47Z

2

The logic here is wrong:

  while((c = getchar()) != '\n') {
      realloc(input, (sizeof(char)));
      input[i++] = c;
  }

You're not actually increasing the size of the buffer, and you are also discarding the result of realloc.

Try:

  while ((c = getchar()) != '\n') {
    // Note: you need one extra character for the terminator, so for the
    // first char, when `i` is 0, then you need room for two `char`s in
    // the buffer - one for the first input character and one for the
    // terminator. And so on...
    char * temp = realloc(input, i + 2); // NB: realloc can, and sometimes does, fail
    if (temp == NULL) // if realloc failed then exit program
        exit(1);
    input = temp;     // otherwise update input...
    input[i++] = c;
  }

Also, since you are always going to be calling realloc on every character (which is very inefficient, incidentally, but it works), this line:

input = (char *) malloc(sizeof(char));

(which should not have a cast, BTW, since this is C, not C++) can just be:

input = NULL;

And one final bug:

char c;

should be:

int c;

otherwise your while loop may never terminate, since EOF can only be properly represented as an int.

So the final fixed program should look something like this:

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    int i = 0;
    int c;
    char * input = NULL;

    printf("Input a string, press ENTER when done: ");

    while ((c = getchar()) != '\n') {
        // Note: you need one extra character for the terminator, so for the
        // first char, when `i` is 0, then you need room for two `char`s in
        // the buffer - one for the first input character and one for the
        // terminator. And so on...
        char * temp = realloc(input, i + 2); // NB: realloc can, and sometimes does, fail
        if (temp == NULL) // if realloc failed then exit program
            exit(1);
        input = temp;     // otherwise update input...
        input[i++] = c;
    }

    input[i] = '\0';
    printf("\nYou've entered the string: %s\n", input);

    return 0;
}

edited Apr 11, 2017 at 21:12

answered Apr 11, 2017 at 20:55

Paul R

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5 Comments

Amitrajit Bose Over a year ago

That is great sir. But I am still getting a TLE. After replacing this snippet while((c = getchar()) != '\n') { _ realloc(input, (sizeof(char)));_ _ input[i++] = c;_ } with yours.

Paul R Over a year ago

Did you apply all THREE bug fixes above ?

Amitrajit Bose Over a year ago

No basically I applied only the one in the while loop. Should I apply all of them?

Paul R Over a year ago

Um, think about it a minute - you have (at least) three bugs in your code - should you just fix one of the bugs, or should you fix all three ? (I've now added a complete, fixed, version of the code, above, for your convenience.)

Amitrajit Bose Over a year ago

Well. All three. :D Yea. Now it's running smooth. Thank You Sir.

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