How to initialize rows of 2D array of strings in C

Question

I want to store strings in a 2D array using pointers but I'm confused with how to do it. The examples I've seen use only arrays of ints or use the brackets[] to allocate a fixed size of memory. So I'm trying to initialize my 2D array of strings and this is what I have:

char ** stringArr = (char**)malloc(/*I don't know what goes here*/);
for(i = 0; i < rows; i++)
    stringArr[i] = (char*)malloc(cols *sizeof(char));

As you can see the parameter for my first call of malloc, I am stuck as to what to put there if I want an exact x number of rows, where each row stores a string of chars. Any help would be appreciated!

John Zwinck · Accepted Answer · 2015-09-29 02:31:02Z

2

Do this, because you're allocating some number of pointers:

malloc(rows * sizeof(char*))

answered Sep 29, 2015 at 2:31

John Zwinck

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Corb3nik · Accepted Answer · 2015-09-29 02:31:50Z

0

You will want to use the number of rows.

char ** stringArr = malloc(rows * sizeof(char*));

Also, do not typecast the return value of a malloc() call.

answered Sep 29, 2015 at 2:31

Corb3nik

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2 Comments

David Velasquez Over a year ago

Thanks for your reply but why shouldn't I typecast malloc()? Even cplusplus.com typecasts it in its reference page on malloc().

Corb3nik Over a year ago

@Ghost_Stark Here is a protected link on the subject! Hope it makes things clearer : stackoverflow.com/questions/605845/…

chux · Accepted Answer · 2015-09-29 02:44:04Z

0

Use sizeof *ptr * N.

Notice how this method uses the correct type even if stringArr was char ** stringArr or int ** stringArr.

stringArr = malloc(sizeof *stringArr * rows);
for(i = 0; i < rows; i++)
    stringArr[i] = malloc(sizeof *(stringArr[i]) * cols);

answered Sep 29, 2015 at 2:44

chux

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