getting the length of an array after passing as a char*

Question

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Sizeof an array in the C programming language?

I have code of the following form:

void arrLen(char* array){
    // print length
}

int main(){
    char bytes[16];
    arrLen(bytes);
}

What I need to do is get the length of the underlying array and print it out. The issue is that the array MAY CONTAIN NULL CHARACTERS, and therefore strlen would not work. is there any way to do this, or would I have to pass it as a char[] instead of a char*?

Well that's not really how arrays are handled in C. They are not actual objects that carry length information. If you have access to C++11, use std::array. If not, you must keep track of the memory length yourself at all times — im so confused, Jan 04 '13 at 14:43
@AK4749 Thank, but I do not have access to c++11, unfortunately. Is there a boost module that could perform this? — ewok, Jan 04 '13 at 14:44
@ewok indeed: http://www.boost.org/doc/libs/1_52_0/doc/html/array.html — im so confused, Jan 04 '13 at 14:45
Does the char[] end in a null character ('\o')? You could use that to find the end of the array. — Raymond Saltrelli, Jan 04 '13 at 14:49
@RaySaltrelli well seeing as he says the char[] might be sprinkled with extra '\0's, I don't quite see how you can determine which is "the last" '\0' — im so confused, Jan 04 '13 at 14:50
@ewok: Also check if you have access to TR1, there is `std::tr1::array`. Boost also provides wrappers which automatically provide some version http://www.boost.org/doc/libs/1_52_0/doc/html/boost_tr1.html. — Benjamin Bannier, Jan 04 '13 at 14:51
^ Forgot about TR1 array. good call. Very few people don't have TR1 now — im so confused, Jan 04 '13 at 14:52

undu · Accepted Answer · 2013-01-04T15:00:17.320

C style arrays do not carry length information with them.

You have several solutions avaliable :

pass the size of your array to your function :

void arrLen(char* array, std::size_t size)

use std::array if you have access to C++ 11 (or TR1 as mentionned in comments)

 std::array<char, 16> bytes;

If you don't have access to std::array, you can use boost::array. They are roughly equivalent to the standard one.

Regarding your comment

would I have to pass it as a char[] instead of a char*?

Arrays are always passed to functions as pointers, so void arrLen(char* array) and void arrLen(char[] array) are equivalent.

score 3 · Answer 2 · edited Jan 04 '13 at 15:58

3

or would I have to pass it as a char[] instead of a char*?

Changing the declaration of arrLen() to take a parameter of type char[] instead of char* would have no effect, as in this scenario they are equivalent.

A possible solution would be to use a template function:

#include <iostream>

template <typename T, int N>
void array_size(T (&a)[N])
{
    std::cout << N << "\n";
}
int main()
{
    array_size("hello");
    char bytes[16];
    array_size(bytes);
    return 0;
}

See http://ideone.com/l7GxDp for demo.

Alternatives to using an array:

std::string; these can contain NULL characters:
```
    std::string s("hello\0world", 11);
```
std::vector<char>
boost::array<char, 16> (available as std::array<char,16> in C++11, or std::tr1::array in most every C++ compiler on the market)

edited Jan 04 '13 at 15:58

Yakk - Adam Nevraumont

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answered Jan 04 '13 at 14:46

hmjd

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Considering that `std::array` is standard as of C++11, I'd prefer seeing a reference to that as well when mentioning boost::array. – Agentlien Jan 04 '13 at 14:55
@Agentlien, I know that but in a comment to the question the OP states that C++11 is not available. – hmjd Jan 04 '13 at 14:57
Sure, but I'd still add a comment, possibly within parentheses. As such: *boost::array (available as std::array in C++11* That would help people who come to this question later and do not have that restriction. – Agentlien Jan 04 '13 at 15:08

score 2 · Answer 3 · answered Jan 04 '13 at 14:44

2

In the traditional way (C) one has to pass the size as an attribute of the function

void arrLen(char* array, int maxSize){
    // print length
}

int main(){
    char bytes[16];
    arrLen(bytes, 16);
}

answered Jan 04 '13 at 14:44

Helio Santos

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score 1 · Answer 4 · answered Jan 04 '13 at 14:45

1

arrays in C/C++ aren't handled that way per say, you have to manually keep track of the length yourself.. change your function to something like void arrLen(char * array, int size) and specify the size whenever you call the function.

answered Jan 04 '13 at 14:45

iKlsR

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score 1 · Answer 5 · answered Jan 04 '13 at 14:47

1

If you have a C++11, may I suggest that you use std::array, it keeps track of it size.

answered Jan 04 '13 at 14:47

Caesar

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score 1 · Answer 6 · answered Jan 04 '13 at 14:48

Unfortunately, char[] and char* parameter types are equivalent.

You have two options:

Pass the length along with the pointer like one would in C:

void arrLen(char* array, size_t length)

or use a template:

template<size_t sz>
void arrLen(char (&array)[sz])

getting the length of an array after passing as a char*

6 Answers6