Try to use in malloc (c)

Question

I try to create arr[N][N] and I wrote this:

int **arr;
arr = (int**)malloc((strlen(str) - 1) * sizeof(int*));
for (i = 0; i < strlen(str) - 1; i++) {
    arr[i] = (int*)malloc((strlen(str) - 1) * sizeof(int));
}

If in to example the size of str is 6 I need to create arr[5][5], how can I fix it?

What problem are you facing? Also, in the first malloc, `int*` should be replaced with `int`. — Harshil Sharma, Dec 01 '13 at 13:27
Do you really want the width and length if you 2D array to be the same? — Fiddling Bits, Dec 01 '13 at 13:35
I think maybe you want the size of the array to be `strlen(str)` and not `strlen(str) - 1`, but otherwise the code looks correct. What problem is it giving you? — Tim Pierce, Dec 01 '13 at 13:35
the problem is that when I try to get the arr[0][20] in to example its work,the arr isn't in the size that I want — user2976686, Dec 01 '13 at 14:03

Bruce Dean · Answer 1 · 2013-12-01T14:38:31.137

2

Why wouldn't you just do this (N could be 5):

int **arr;
arr = malloc(N*sizeof(int*));
for (int i = 0; i < N; i++) {
    arr[i] = malloc(N*sizeof(int));
}

edited Dec 01 '13 at 14:38

answered Dec 01 '13 at 13:35

Bruce Dean

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Removed casts from `malloc` calls... in C, `malloc`'s returned pointer shouldn't be cast, at least not in this case – Elias Van Ootegem Dec 01 '13 at 13:55
@EliasVanOotegem, yes your quite right, good discussion here too: http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc. Thanks. – Bruce Dean Dec 01 '13 at 13:59
I got error in the line arr = malloc( N * sizeof(int*) ); thanks – user2976686 Dec 01 '13 at 14:10
@user2976686, Don't forget to include the standard library, which contains malloc. So above main() you should have: #include – Bruce Dean Dec 01 '13 at 14:24
@roybatty,yes I include it – user2976686 Dec 01 '13 at 14:37

chux - Reinstate Monica · Answer 2 · 2013-12-01T18:00:01.053

Use N rather than strlen(str) - 1.

int **arr;
arr = malloc(N * sizeof(*arr));
for (i = 0; i < N; i++) {
    arr[i] = malloc(N * sizeof(arr[i][0]));
}

Usually when strlen(str) comes into play, that allocation size is strlen(str) + 1 (in the for loop only).

From subsequent comments, there is some confusion. Now I assume OP wants to store an array of strings. Assume there are up to N strings.

#include <stdlib.h>
char **arr;
// See note on casting the results of malloc()
arr = (char **) malloc(N * sizeof(*arr));
for (i = 0; i < N; i++) {
  char str[100];
  scanf("%99s", str);
  size_t siz = strlen(str) + 1;
  arr[i] = (char *) malloc(siz);
  memcpy(arr[i], buf, siz);
}

Checking the results of malloc() is a good idea for robust code as in

arr = (char **) malloc(N * sizeof(*arr));
if (arr == NULL) handle_error();  // maybe exit?

Casting note: OP reports twice problems with code that uses a cast-less malloc(). Not having the cast, with compliant compilers it should not be a problem. But if OP is using a C++ compiler or something non-compliant, casting should fix that issue. With modern C, casting malloc() results it is tolerable, but frowned upon.

Why have this in the loop: `N * sizeof(arr[i][0])`? Seems like a waste of CPU cycles to recalculate it every loop iteration. — Fiddling Bits, Dec 01 '13 at 13:40
@Bit Fiddling Code Monkey I take it you are suggesting to create the `size_t len = N * sizeof(arr[i][0]);` before the loop and use `arr[i] = malloc(len);` inside the loop. Should `N` be a constant, the posted code I'd expect a bit faster. Should `N` be a variable, your idea would be faster. In either case, profiling would provide definite results. I'd expect margin differences. — chux - Reinstate Monica, Dec 01 '13 at 13:55
hi,first thanks!..this is write me error in the line: arr = malloc(N * sizeof(*arr)); — user2976686, Dec 01 '13 at 14:09

Try to use in malloc (c)

2 Answers2