Copy from 2d dynamic char array to another 2d dynamic array in c

Question

I wanted to copy the content of a two dimensional array into another two dimensional char array. I used the following for loop with memcpy but it is not working as desired. So I have two questions.

What is wrong with this code? and

Is there a way to do it without use of iteration?

for (int i = 0; i < count; i++) {
    memcpy(&buf_copy[i], buf[i], sizeof(buf[i]));
}

Both buf and buf_copy are 2d dynamic char arrays.

Edit: declarations of the arrays

char **buf;
char **buf_copy;

EDIT 2: Here is how memory is allocated to them

void intit_buf()
{
    buf = (char**)malloc(BUFFER * sizeof(*buf));
    for (int i = 0; i < BUFFER; i++)
        buf[i] = (char*)malloc(sizeof(char) * 33);

    //initialize buf_copy
    buf_copy = (char**)malloc(BUFFER * sizeof(*buf_copy));
    for (int i = 0; i < BUFFER; i++)
        buf_copy[i] = (char*)malloc(sizeof(char) * 33);
}

*"Both buf and buf_copy are 2d dynamic char arrays"* - I have a strong suspicion they are not. That there is pointer business going on, and you fail because you don't understand what `sizeof` would do in that case. — StoryTeller - Unslander Monica, Dec 04 '17 at 08:24
When you say "2d dynamic char arrays" do you mean `char **`? Then you have multiple problems. Please create a [Minimal, **Complete**, and Verifiable Example](http://stackoverflow.com/help/mcve) to show us. — Some programmer dude, Dec 04 '17 at 08:24
I see. Then I'm tempted to offer this is a duplicate https://stackoverflow.com/questions/42094465/correctly-allocating-multi-dimensional-arrays — StoryTeller - Unslander Monica, Dec 04 '17 at 08:27
@John_D show as that code too. Post a minimal example *complete* example please. — gsamaras, Dec 04 '17 at 08:28
`sizeof(buf[i])` will be the size of a pointer, not the length of the string or the size of the memory in `buf[i]`. — Paul Ogilvie, Dec 04 '17 at 08:31
@PaulOgilvie comment answer and the answer below answers my first question. How about my second question? Is it possible to do it without iteration? — John_D, Dec 04 '17 at 08:51
@John_D I updated the answer to answer your second question too. — gsamaras, Dec 04 '17 at 08:54
No, it cannot be done without ireation because `malloc` may not have returned contigeous memory. — Paul Ogilvie, Dec 04 '17 at 08:57
WHAT IS WITH THE DOWNVOTE? Nobody can learn without making mistakes!! — John_D, Dec 04 '17 at 09:19
See [Correctly allocating multi-dimensional arrays](https://stackoverflow.com/questions/42094465/correctly-allocating-multi-dimensional-arrays), it explains why memcpy doesn't work, why your current code is _not_ a 2D array, and what you can use instead. — Lundin, Dec 04 '17 at 09:33
John, although I did not downvote your question, your erroneous use of `sizeof`, thinking it will return the size of dynamically allocated memory, is worthy of a downvote (you didn't read the book). — Paul Ogilvie, Dec 04 '17 at 09:36

gsamaras · Accepted Answer · 2017-12-04T09:02:03.070

1

What is wrong with this code?

The first and last parameters of the function call.

Change this:

memcpy(&buf_copy[i], buf[i], sizeof(buf[i]));

to this:

memcpy(buf_copy[i], buf[i], 33);

since you need to give the size of the column, not the size of a pointer!

Moreover, notice that you need to pass buf_copy, just like you did for buf, without taking the address of it or something, since memcpy() expects pointers for its two first parameters, and buf_copy[i] and buf[i] are pointers already.

Is there a way to do it without use of iteration?

It cannot be done without ireation because malloc() may not have returned contigeous memory. Read more in C / C++ How to copy a multidimensional char array without nested loops?

PS: Unrelated to your problem, but in general: Do I cast the result of malloc? No!

edited Dec 04 '17 at 09:02

answered Dec 04 '17 at 08:39

gsamaras

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What do you think about my second question though? – John_D Dec 04 '17 at 08:46
Now i can accept your answer. Earlier I didn't want my question marked "answered" soon because I was waiting for reply regarding my second question. And the dimension of the array is known at compile time. Thanks for the link – John_D Dec 04 '17 at 09:00
More general, as soon as we talk about an array of pointers to rows which are dynamicallt allocated, it cannot be done without iteration. `char a[10][20]` is a contigeous block of memory, so is `char b[n][m]`. But `char **c` with `c=malloc(n*sizeof(char *))` will not result in a 2-dimensional array of contigeous memory. – Paul Ogilvie Dec 04 '17 at 09:07
Right, @PaulOgilvie thanks for helping me improve the answer! – gsamaras Dec 04 '17 at 09:14

Copy from 2d dynamic char array to another 2d dynamic array in c

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