How to get the length of an array in C? Is "sizeof" a solution? [duplicate]

Question

Possible Duplicate:
Sizeof an array in the C programming language?
Why does a C-Array have a wrong sizeof() value when it's passed to a function?

See the below code and suggest me that what is the difference of "sizeof" keyword when I used like this:

#include<stdio.h> #include<conio.h> void show(int ar[]); void main() {     int arr[]={1,2,3,4,5};     clrscr();     printf("Length: %d\n",sizeof(arr));     printf("Length: %d\n",sizeof(arr)/sizeof(int));     show(arr);     getch(); } void show(int ar[]) {    printf("Length: %d", sizeof(ar));    printf("Length: %d", sizeof(ar)/sizeof(int)); } 

But the output is like this:

Output is:

Length: 10

Length: 5

Length: 2

Length: 1

why I am getting like this; If I want to take the entire data from one array to another array the how can I do?

Suggest me If anyone knows.

Answer 1

Arrays decay to pointers in function calls. It's not possible to compute the size of an array which is only represented as a pointer in any way, including using sizeof.

You must add an explicit argument:

void show(int *data, size_t count); 

In the call, you can use sizeof to compute the number of elements, for actual arrays:

int arr[] = { 1,2,3,4,5 };  show(arr, sizeof arr / sizeof *arr); 

Note that sizeof gives you the size in units of char, which is why the division by what is essentially sizeof (int) is needed, or you'd get a way too high value.

Also note, as a point of interest and cleanliness, that sizeof is not a function. The parentheses are only needed when the argument is a type name, since the argument then is a cast-like expression (e.g. sizeof (int)). You can often get away without naming actual types, by doing sizeof on data instead.

Answer 2

Thats the reason why, when writing a function that takes an array, two parameters are declared. one that is a pointer to the array, the other that defines the size of the array.