Initializing dynamic character array in C [duplicate]

Question

I'm making a function that searches through a sentence to find a letter. Right now, I'm having trouble initializing a dynamic array for the user's sentence.

char * get_info(char *ch) {
  char *str;
  int i = 0;

  printf("Enter a sentence to seach: ");
  while ((str[i++] = getchar()) != '\n');
  str[i] = '\0';
  printf("Enter a character to search for: ");
  *ch = getchar();

return str;
}

The problem lies when I call str[i++]. Nothing I've tried or looked has helped so far. Any suggestions? Thanks!

So, what exactly happened when you called str[i++], and what did you expect to happen instead? Also, please post the remainder of the program. — zwol
– zwol, Commented Jun 13, 2014 at 2:37
@self. -- no, that is a different, more complicated situation (array of arrays). — zwol
– zwol, Commented Jun 13, 2014 at 2:38
@Zack The principle is the same, there are also dozens of existing questions that solve this. Find one that you think is adequate and mark as duplicate. — this
– this, Commented Jun 13, 2014 at 2:41

jxh · Accepted Answer · 2014-06-13 03:33:43Z

3

The problem is you are dereferencing an uninitialized variable. You need to allocate memory for str somehow.

Fortunately for you, there is a function that does this for you, called getline().

char * get_info(char *ch) {
  char *str;

  printf("Enter a sentence to seach: ");
  str = NULL;
  getline(&str, NULL, stdin);
  printf("Enter a character to search for: ");
  *ch = getchar();

  return str;
}

edited Jun 13, 2014 at 3:33

answered Jun 13, 2014 at 2:38

jxh

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3 Comments

zwol Over a year ago

... I should have seen that. +1 for you.

derekerdmann Over a year ago

How is not guarding against a buffer overrun the fundamental problem when there isn't even a buffer allocated? Filling the input string using getchar() and dynamically using a larger buffer as needed essentially mitigates that problem.

derekerdmann Over a year ago

@jxh I can see that, but that's because it isn't doing memory management at all. The OP is asking to dynamically read in the string, so it will need to be expanded at some point once the original buffer's limit is reached. getline() doesn't change that.

thurizas · Accepted Answer · 2014-06-13 03:18:32Z

the declaration char* str, can be read as str is a variable that will hold a pointer to a character. No where in your code do you actually allocate this memory for a character (or an array of characters)

If you were to do the following:

char *str;
str = malloc(15*sizeof(char));

You have not only declared that str will contain a pointer to a character (or an array of characters), but also allocated this memory.

Now while this solves your original question, as jxh points out, this introduces another error. This code suffers from a classical heap-based buffer overflow, in that if I supply an input line with, say, 30 characters I will use all the memory allocated to the array and over-write other data.

Additionally, you do need to guard against memory allocation failure and attempt to gracefully handle it.

Giancarlo Manuel Guerra Salvá · Accepted Answer · 2014-06-13 03:38:18Z

0

You could just implement something like this, if your goal is to find a particular letter within a sentence. It's certainly not perfect but should help you get an idea. Cheers!

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

int main(){

    char sentence[30]; //store your sentence

    char letter[3]; //here we store the letter along with the EOL(\0)
    int i;


    puts("Please enter a sentence to search through: ");
    fgets(sentence, 30, stdin); //
    puts("Now enter the letter: ");
    scanf("%2s",letter); //max size for input is 2 including the EOL 


    if(strstr(sentence, letter)){

        printf("The letter '%s' can be found in the sentence.\n",letter);
    }

}

answered Jun 13, 2014 at 3:38

Giancarlo Manuel Guerra Salvá

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