How to get array size within function?

Question

I am having trouble finding a way to get the array size from within a function. Here is my code:

#include <stdio.h>

void printBuff(char *buf);
int main()
{
    char arr[12] = "csdlnclskjn";
    printf("Array size: %d, element size: %d. ",sizeof(arr), sizeof(arr[0]));
    printBuff(arr);
    return 0;
}

void printBuff(char *buf){
    printf("Array size: %d, element size: %d.",sizeof(buf), sizeof(buf[0]));
}

As seen above, printBuff does the same as the second line in the main function. However, the outputs are different:

Array size: 12, element size: 1. Array size: 4, element size: 1.

Thinking about it, I understand why the output is 4 in the printBuff() method. In fact, arr is a pointer to the first element of the array. On a 32-bit architecture, sizeof(arr) will then return 4, on 64-bit one it will return 8. What I do not understand is why sizeof(arr) returns the size of the array instead of the number of bytes of the pointer when used in the main() function. After all, arr, when invoked inside main(), is still a pointer, right?

So my questions are:

How come sizeoff() is interpreted differently depending on the context in which it is used? What does this depend on?
How to get the actual array size (number of elements in array) from within a function, without passing the size as an argument, without using methods such as iterating over the array while incrementing a counter until '\0' is reached - just the simplest way to get array size regardless of the context.
Incidentally, where does the compiler/system responsible for remembering the size of the array store the size of the array? How is it associated with the array and how is it retrieved?

I wanted to iterate though an array using sizeof(buf), / sizeof(buf[0]) as the size of the array but apparently that is not possible.

Your array decays to a pointer in the function. That is all that is known so you have to pass the array length too. the size of an array (in main) is not the same as the size of a pointer (in the function). — Weather Vane
– Weather Vane, Commented Sep 10, 2016 at 19:20
@WeatherVane: It's not converted in the function. It's converted when the argument expression is evaluated, before the function is called. — Keith Thompson
– Keith Thompson, Commented Sep 10, 2016 at 19:50
Possible duplicate of How to get the real and total length of char * (char array)? — Elias Van Ootegem
– Elias Van Ootegem, Commented Sep 10, 2016 at 19:54
@WeatherVane: It's an important distinction. There's a common misconception that passing an array expression to a function is a special case. It isn't. The "conversion" happens when the array expression is evaluated (with a few stated exceptions). — Keith Thompson
– Keith Thompson, Commented Sep 10, 2016 at 19:59

Community · Accepted Answer · 2017-05-23 10:29:14Z

3

so here are my answers for your questions:

The array is "converted" into char* type when passed into the function (with the char* parameter).
AFAIK there is no such way. You could use strlen function for strings. Otherwise, you have to pass the length as parameter.
See How does an array pointer store its size?

Don't use sizeof(buf)/sizeof(buf[0]) to get length of an array when passing array parameters. See this. for more information.

edited May 23, 2017 at 10:29

CommunityBot

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answered Sep 10, 2016 at 19:42

user5012597

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2 Comments

Keith Thompson Over a year ago

The argument against sizeof(buf)/sizeof(buf[0]) applies only to array parameters (because there's no such thing as an array parameter in C). It doesn't work because what looks like an array parameter is really a pointer parameter. sizeof buf / sizeof buf[0] works correctly if buf is an array object.

user5012597 Over a year ago

Thank you for your comment. I agree and I am editing my answer. Although, it is described in the link I have included.

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