Lets say I have an array like
int arr[10][10];
Now i want to initialize all elements of this array to 0. How can I do this without loops or specifying each element?
Please note that this question if for C
1
-
Duplicate: stackoverflow.com/questions/201101/…ptomato– ptomato2010-05-23 13:49:18 +00:00Commented May 23, 2010 at 13:49
Add a comment
|
7 Answers
The quick-n-dirty solution:
int arr[10][10] = { 0 };
If you initialise any element of the array, C will default-initialise any element that you don't explicitly specify. So the above code initialises the first element to zero, and C sets all the other elements to zero.
5 Comments
Laz
This will initialize all elements to 0? Why do you say its dirty?
Mahmoud Al-Qudsi
Just to clarify: It doesn't "follow suit," it forces them to zero. If you wrote
= { 1 }; it would put the first value as 1, and the rest would still be zeros.
Marcelo Cantos
I say quick-n-dirty because, even though it is logically correct and guaranteed to work, it may confuse maintainers. I consider it a shortcut, and avoid using it unless the array or data structure is too large to conveniently initialise every element.
Marcelo Cantos
Good point @Computer Guru; I didn't consider the ambiguity in my statement. I've amended it.
R.. GitHub STOP HELPING ICE
+1 for pointing out the little-known
{0}. It should be noted that the {0} initializer works for any type in the C language - integer types, floating point types, pointer types, array types (of any type), structures, unions, enums, you name it.Besides the initialization syntax, you can always memset(arr, 0, sizeof(int)*10*10)
6 Comments
dmckee --- ex-moderator kitten
Note that this still take O(N) time for N elements. On the other hand it is probably a faster O(N) than the one you code by hand (or at least no slower).
SoapBox
memset uses a loop, so this doesn't answer the question.
Alok Singhal
memset might also be wrong for pointer or floating-point arrays.
Terry Mahaffey
Yes, it's technically O(n). It's unavoidable. The only way to set an arbitrary amount of bytes to the same value in constant time is with a very large magnet.
dmckee --- ex-moderator kitten
@Terry: Yes, unavoidable. I wasn't criticising your answer (indeed, you got my vote), but rather trying to prevent Shlemiel-the-painter problems for users of "high level" language users who may have understood this action as being a "single call" (I have see evidence of this issue on other posts on SO).
|
You're in luck: with 0, it's possible.
memset(arr, 0, 10 * 10 * sizeof(int));
You cannot do this with another value than 0, because memset works on bytes, not on ints. But an int that's all 0 bytes will always have the value 0.
2 Comments
Thomas
... initializing all elements to -1. Good point, hadn't thought of that.
R.. GitHub STOP HELPING ICE
You can do it with any other value which, as bytes, consists of the same value in each byte.
n*(UINT_MAX+1ULL)/255 is the family of such values (n=0,1,...,UCHAR_MAX).int arr[10][10] = {0}; // only in the case of 0
Comments
int myArray[2][2] = {};
You don't need to even write the zero explicitly.
2 Comments
Mahmoud Al-Qudsi
Yeah, sorry about that. Updated.
avakar
I think in C you have to write the zero (the empty braces are, however, valid C++).
Defining a array globally will also initialize with 0.
#include<iostream>
using namespace std;
int ar[1000];
int main(){
return 0;
}
1 Comment
Matthew Cole
OP asked for C, this is C++