I created a dynamic array ,and i need to initialize all the members to 0. How can this be done in C?
int* array;
array = (int*) malloc(n*sizeof(int));
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5 Answers
In this case you would use calloc():
array = (int*) calloc(n, sizeof(int));
It's safe to assume that all systems now have all zero bits as the representation for zero.
§6.2.6.2 guarantees this to work:
For any integer type, the object representation where all the bits are zero shall be a representation of the value zero in that type.
It's also possible to do a combination of malloc() + memset(), but for reasons discussed in the comments of this answer, it is likely to be more efficient to use calloc().
8 Comments
Matteo Italia
It's more than "safe to assume", I'd say that it's actually guaranteed by §6.2.6.2.
ouah
@Mysticial it is guaranteed by the standard (C99, 6.2.6.2p5) "For any integer type, the object representation where all the bits are zero shall be a representation of the value zero in that type."
Mysticial
Yes, you are right, I found the line that says this. Adding to my answer.
Matteo Italia
Value bits => "Each bit that is a value bit shall have the same value as the same bit in the object representation of the corresponding unsigned type"; Sign bits => "If the sign bit is zero, it shall not affect the resulting value." Padding bits aren't relevant.
Mat
The cast is unnecessary too in C.
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memset(array, 0, n*sizeof(int));
Or alternatively, you could allocate your block of memory using calloc, which does this for you:
array = calloc(n, sizeof(int));
calloc documentation:
void *calloc(size_t nmemb, size_t size);
The calloc() function allocates memory for an array of nmemb elements of size bytes each and returns a pointer to the allocated memory. The memory is set to zero. ...
Comments
Use calloc function (usage example):
int *array = calloc(n, sizeof(int));
From calloc reference page:
void *calloc(size_t nelem, size_t elsize);The
calloc()function shall allocate unused space for an array ofnelemelements each of whose size in bytes iselsize. The space shall be initialized to all bits 0.
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memset(array, 0, n * sizeof *array);
Comments
for (i=0; i<x; ++i){
array[i]=0;
}
That should do the trick. I guess you have to make a loop that makes 0 every element in the array. Memset could also work for you.
2 Comments
Doncho Gunchev
Hey, you do not have to write a program that is interactively proving the answer correct. Finding the one line that does the trick in this source is way harder...
BugShotGG
@dgunchev Yes you are right but i was just writing something similar in c arm when i saw the question by chance and thought about a fast answer. :)