How return "int" value in a template c++, with strings as parameters

Question

I'm learning c++ for the first time. I want to develop a template to calculate length of a const char*, char[] without having to redefine the same function.

I have tried,

template <class N> N length(N str){
    int pos = 0;
    while (str[pos] != '\0'){pos++;}
    return pos; 
}

But I got an error, since I should return N class. Can I fix this?

FWIW, both a pointer and an array argument will work with a const char* parameter. No need for a template there. There's also std::string and std::string_view, which have functions to get their length (among many other useful features). — Qaz
– Qaz, Commented Oct 12, 2020 at 4:02
You can just make your function return an int, it doesn't have to return an N just because it's templated over class N. — Nathan Pierson
– Nathan Pierson, Commented Oct 12, 2020 at 4:03
If str is a const char*, what should the return type be? If str is a std::string instance, what should the return type be? Do you see a pattern? — Karl Knechtel
– Karl Knechtel, Commented Oct 12, 2020 at 4:10

eerorika · Accepted Answer · 2020-10-12 04:03:48Z

2

I want to develop a template to calculate length of a const char*, char[] without having to redefine the same function.

char[] will implicitly convert to a const char*, so you don't need a template. It is sufficient to write a function that accepts a const char*. Also, such function exists in the standard library so there is no need to write the function except for exercise purposes.

But I got an error, since I should return N class.

Your return statement has the type int, which makes sense for returning a length. But you've declared that the function returns N object instead. This does not make sense for a function that is supposed to return a length. Solution: Declare the function to return an integer type. std::size_t is conventionally used for array / string lengths.

answered Oct 12, 2020 at 4:03

eerorika

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Comments

asmmo · Accepted Answer · 2020-10-12 04:42:38Z

1

There are functions to do that but if you want to use templates, here is a function.

Note that u return N and u ask how to solve this problem while u should return size_t.

#include <iostream>

template <size_t N>
constexpr size_t length(const char (&arr) [N]){
    return N-1;
}
int main(){
    char arr [3] = "me";
    std::cout << length(arr) << " " << length("me");
}

Demo

Note that this template won't work if \0 is not at the end of the characters array and won't work with char* as @errorikta has mentioned.

It's just for practice. to know the size of those strings use std::strlen from <cstring> and you can see a possible implementation in the attached link.

edited Oct 12, 2020 at 4:42

answered Oct 12, 2020 at 4:15

asmmo

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1 Comment

eerorika Over a year ago

What if the array is char arr [42] = "me";? You might also want to clarify that this template does not work with a pointer to char.

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