How to find the size of an integer array in C.
Any method available without traversing the whole array once, to find out the size of the array..
Thankss..
No - in many cases this will *not* work.
Paul R
2010-05-05 13:01:09
It works, however `site_t len = sizeof(array)/sizeof(array[0]);` it's a bit better (i.e. it still works when datatype of array elements has been changed.
Grzegorz Gierlik
2010-05-05 13:02:02
@Grzegorz: Show us how it works for this array: `void f(int array[]) { site_t len = sizeof(array)/sizeof(array[0]); }`
sbi
2010-05-05 13:03:54
@Paul R: Why not? Unless `array` is an incomplete type (one case); if it's an array of `int` this will work.
Charles Bailey
2010-05-05 13:04:05
@sbi: The poorly named `array` parameter is actually a pointer.
caf
2010-05-05 13:16:14
@caf: Yes, and that is something beginners keep running into. (I suppose a question like this _is_ from a beginner.)
sbi
2010-05-05 13:20:52
@Charles: see sbi's answer as to why this only works for certain cases.
Paul R
2010-05-05 13:26:48
I didn't see sbi's first comment when I wrote mine. Yes, in C99 variable length arrays are a second case. Obviously if `array` isn't an array but is actually a pointer then it's not going to work but there was no indication that this might be the case in the answer.
Charles Bailey
2010-05-05 13:42:40
+13
A:
If the array is a global, static, or automatic variable (int array[10];), then sizeof(array)/sizeof(array[0]) works.
If it is a dynamically allocated array (int* array = malloc(sizeof(int)*10); or void f(int array[])), then you cannot find its size at run-time. You will have to store the size somewhere.
Note that sizeof(array)/sizeof(array[0]) compiles just fine even for the second case, but it will silently produce the wrong result.
sbi
2010-05-05 12:58:32