CODEWITHSUNDEEP
c
user379888
user379888

Reputation:

memory allocation for array of pointers

I need to allocate memory for a pointer which needs to be used as a 2d array.I know how to allocate memory for char pointers and int pointers I am confused how memory is allocated of a array of pointers.A pictorial representation of the reason would be very helpful,also is the code below fine?

char *names[5];
for(i=0;i<5;i++)
{
 names[i]=(*char)malloc(sizeof(char));
}

Upvotes: 5

Views: 40227

Answers (4)

rmatos
rmatos

Reputation: 11

Emphisizing what Jack said in in the end: his code works only if the array is declared as a pointer-of-pointers by using **.

When the array is declared as

char *names[5];

I guess the right way of allocating memory is almost as Ak1to did, but multiplying by the wanted size of the string of chars:

char *names[5];
for(i=0;i<5;i++)
{
  names[i]=(char *)malloc(sizeof(char)*80);
}

else the compiler throws an error about incompatible types.

Upvotes: 0

v1Axvw
v1Axvw

Reputation: 3054

What you're doing is allocating space for 5 chars. You could write this and it'll have the same result:

char *names = (char *)malloc(sizeof(char) * 5);

If you want to have an array of pointers, I think this'd be the best code

char **names = (char **)malloc(sizeof(char *) * 5);

I'm not a super-coder, but as what I now, this is the right solution.

Upvotes: 3

kuriouscoder
kuriouscoder

Reputation: 5582

Also, in short you can do the following too in this case for allocating storage for five characters

names = (char*)malloc(5 * sizeof(char))

Upvotes: 0

Jack
Jack

Reputation: 133577

No, this is not because you are allocating the array assuming a dimension of just 1 element of primitive type char (which is 1 byte).

I'm assuming you want to allocate 5 pointers to strings inside names, but just pointers.

You should allocate it according to the size of the pointer multiplied by the number of elements:

char **names = malloc(sizeof(char*)*5);

You don't need to allocate them one by one with a loop. Note that you need to specify that it is a pointer-of-pointers by using **

Upvotes: 14

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