Finding length of array inside a function [duplicate]

Question

In the program below the length of the array ar is correct in main but in temp it shows the length of the pointer to ar which on my computer is 2 (in units of sizeof(int)).

#include <stdio.h>  void temp(int ar[]) // this could also be declared as `int *ar` {     printf("%d\n", (int) sizeof(ar)/sizeof(int)); }  int main(void) {     int ar[]={1,2,3};     printf("%d\n", (int) sizeof(ar)/sizeof(int));     temp(ar);     return 0; } 

I wanted to know how I should define the function so the length of the array is read correctly in the function.

Answer 1

There is no 'built-in' way to determine the length inside the function. However you pass arr, sizeof(arr) will always return the pointer size. So the best way is to pass the number of elements as a seperate argument. Alternatively you could have a special value like 0 or -1 that indicates the end (like it is \0 in strings, which are just char []). But then of course the 'logical' array size was sizeof(arr)/sizeof(int) - 1

Answer 2

Don't use a function, use a macro for this:

//Adapted from K&R, p.135 of edition 2. #define arrayLength(array) (sizeof((array))/sizeof((array)[0]))  int main(void) {     int ar[]={1,2,3};     printf("%d\n", arrayLength(ar));     return 0; } 

You still cannot use this macro inside a function like your temp where the array is passed as a parameter for the reasons others have mentioned.

Alternative if you want to pass one data type around is to define a type that has both an array and capacity:

typedef struct {   int *values;   int capacity; } intArray;  void temp(intArray array) {   printf("%d\n", array.capacity); }  int main(void) {     int ar[]= {1, 2, 3};     intArray arr;     arr.values = ar;     arr.capacity = arrayLength(ar);     temp(arr);     return 0; } 

This takes longer to set up, but is useful if you find your self passing it around many many functions.