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.
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14How did you implement this array? In principle either you know the array size in O(1) (known size), O(N) (nil-terminated), or impossible.kennytm– kennytm2010-05-05 12:56:35 +00:00Commented May 5, 2010 at 12:56
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Usually arrays are created as static variable and you must have passed some length while creating it.Jack– Jack2010-05-05 13:35:11 +00:00Commented May 5, 2010 at 13:35
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1@Jack: Why would arrays "usually" be static???sbi– sbi2010-05-05 13:38:01 +00:00Commented May 5, 2010 at 13:38
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int len = strlen(array); ???Joe DF– Joe DF2013-01-18 00:04:18 +00:00Commented Jan 18, 2013 at 0:04
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2"Any method available without traversing the whole array once, to find out the size of the array." - I would rather like know how you would traverse the array without knowing its size beforehand?Christian Rau– Christian Rau2013-05-14 12:05:03 +00:00Commented May 14, 2013 at 12:05
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4 Answers
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 passed as a function argument (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.
4 Comments
Casper Beyer
Perhaps mention and point to a question explaining array decay? Frequent topic
sbi
For the C++ novices arriving at this C question: Everything you ever wanted to know about arrays in C++ in one FAQ.
lolololol ol
Are the parentheses around sizeof necessary?
sbi
@lololololol In principle, you can use
sizeof without the parentheses. Except in some circumstances. I never even tried to read the fineprint, and instead subscribed to the common practice to always employ them.If array is static allocated:
size_t size = sizeof(arr) / sizeof(int);
if array is dynamic allocated(heap):
int *arr = malloc(sizeof(int) * size);
where variable size is a dimension of the arr.
1 Comment
HeavenHM
Dont ever use size name for identifier it may cause problems.
_msize(array) in Windows or malloc_usable_size(array) in Linux should work for the dynamic array
Both are located within malloc.h and both return a size_t
Comments
int len=sizeof(array)/sizeof(int);
Should work.
7 Comments
Grzegorz Gierlik
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.sbi
@Grzegorz: Show us how it works for this array:
void f(int array[]) { site_t len = sizeof(array)/sizeof(array[0]); }
Lara Bailey
@Paul R: Why not? Unless
array is an incomplete type (one case); if it's an array of int this will work.
caf
@sbi: The poorly named
array parameter is actually a pointer.
Paul R
@Charles: see sbi's answer as to why this only works for certain cases.
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