Creating an array of strings using malloc in C

Question

I am completely novice in C, and just learned about the dynamic memory allocation using malloc, realloc, calloc and free.

I want to make a small programm which takes an int number as the number of the strings that will be given and then "scanf" them all. Next play with these strings. For example find the most frequent and print it.
For example when i run the programm and type :
5
car house dog tree tree
It should print :
tree 2

I want scanf-printf because this is the input/output methods i am most familiar with at this point.
My code :

int main (){

int N,i,j ;

char *array;

int *freq;


 scanf("%d",&N);

 array = (char*)calloc(N,sizeof(char*));
 for (i=0;i<=N;i++){    
  scanf( ??? );  
 }

 free(array);  
 return 0;  
}

What should i enter in the scanf fuction in order to properly fill the array with strings ? After i fill it will i use something like strcmp and a for loop in order to scan the array and find the most frequent word ? (I can store the frequencies in the *freq)

Community · Accepted Answer · 2017-05-23 12:03:59Z

3

You want to allocate an array of strings, in other words an array of pointers to characters, and that's exactly what you do allocate. The problem is that you assign the pointer returned by calloc to an array of characters.

You have two choices here actually: Either change your declaration of array to be an "array" pointers to character, e.g. char **array, and then dynamically allocate the individual strings as well. Something like this

// Allocate an array of pointers
char **array = calloc(N, sizeof(*array));

// Allocate and read all strings
for (size_t i = 0; i < N; ++i)
{
    // Allocate 50 characters
    array[i] = malloc(50);  // No need for `sizeof(char)`, it's always 1

    // Read up to 49 characters (to leave space for the string terminator)
    scanf("%49s", array[i]);
}

Or you can change the type of array, to be a pointer to fixed-sized "strings", like this

// Define `my_string_type` as an array of 50 characters
typedef char my_string_type[50];

// Declare a pointer to strings, and allocate it
my_string_type *array = calloc(N, sizeof(*array));

// Read all strings from the user
for (size_t i = 0; i < N; ++i)
{
    // Read up to 49 characters (to leave space for the string terminator)
    scanf("%49s", array[i]);
}

_{Note that I don't cast the result of calloc or malloc. You should never cast void * in C.}

edited May 23, 2017 at 12:03

CommunityBot

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answered Nov 21, 2015 at 12:19

Some programmer dude

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2 Comments

manosgior Over a year ago

Now to the fun part : For the frequency question (the string with the most appearences) should i implement something like this : for (i=0;i<N;++i){ for (j=i+1;j<=N;++j){ result = strcmp (array[i], array [j]) if result =0 freq[i]= freq[i] + 1 } }

Some programmer dude Over a year ago

@ManosG SOmething like that could possibly work, but be careful with the condition for the inner loop.

Pete · Accepted Answer · 2015-11-21 12:20:24Z

1

In the scanf function, you need to select a format and the array you want the data to go. For example:

scanf("%[^\n]", array);

answered Nov 21, 2015 at 12:20

Pete

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Comments

Allen Sun · Accepted Answer · 2015-11-21 14:40:53Z

1

You need to make sure input is no more than the size you applied ,try scanf("%s",array);

answered Nov 21, 2015 at 14:40

Allen Sun

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1 Comment

Allen Sun Over a year ago

I just noticed seems you just make a one-dimensional array but you want to store strings in it ? You need to make a char** to make it .Or it will just be one string with spaces between some chars not between words.

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