Changing a static array to a dynamic one with minimal changes?

Question

I have a C code that creates an array of char pointers as the following:

char* arr[100];

I use each element in this array to point to some string that is being calculated by another function. So basically arr[0] would point to string1 and arr[1] to string2, etc.

This works just fine. However I'm now asked to be more flexible by having the user to specify the number of strings as a parameter.

How can I do this with minimal changes to my code? I understand that I need to use malloc. However I'm getting a lot of warnings in all of the assignment statements I had before. I changed the declaration of the array as the following:

char* arr = (char*)malloc(n * sizeof(char*)); //where n is provided by user

I thought that I only needed to change the declaration. Now all of the assignment statements are giving warnings ("assignment makes integer from pointer without a cast"). The following is an example of an assignment statement:

arr[i] = str; //where str is defined as char* and is calculated by another function

Am I missing something here?

Waynn Lue · Accepted Answer · 2012-03-10T21:17:48.013

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If you're looking to create an array of char *, you need a char **arr. Think of it as an array of char * -- if you had an array of int, you would have int *. Since you have an array of char *s, you need char **.

char** arr = malloc(n * sizeof(char*));

edited Mar 10 '12 at 21:17

answered Mar 08 '12 at 22:51

Waynn Lue

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Please, remove the cast. In C, no cast is needed when converting a `void *` to another pointer. – Danilo Piazzalunga Mar 10 '12 at 17:05
Changed, though I wasn't sure if the OP was using a C or C++ compiler. – Waynn Lue Mar 10 '12 at 21:24

score 2 · Answer 2 · edited May 23 '17 at 11:55

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You are declaring arr as a pointer to char: either a single string or, if you prefer, an array of chars.

To allocate an array of pointers, declare arr as

char **arr = malloc(n * sizeof(char *));

By the way, remove the cast: it is unnecessary in C. See also Question 7.7 in the comp.lang.c FAQ.

edited May 23 '17 at 11:55

Community

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answered Mar 08 '12 at 22:56

Danilo Piazzalunga

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score 1 · Answer 3 · 2012-03-09T00:08:28.550

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you want to declare arr as as char ** since you are pointing to an array of pointers. If you declare arr only as char * (not char ** or char *[]) you only have one "string".

edited Mar 09 '12 at 00:08

answered Mar 08 '12 at 22:53

score 1 · Answer 4 · answered Mar 08 '12 at 22:59

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Don't forget that string is an array too (char *), so u will need array of pointers, which should look like this:

char** arr = (char**)malloc(n * sizeof(char*));

answered Mar 08 '12 at 22:59

Neko

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Changing a static array to a dynamic one with minimal changes?

4 Answers4