Function that will act like an Strlen C++

Question

I want to make a function that will do the same as Strlen does, andI get the error: "string subscript out of range". I tried fixing it but has no idea.

heres the code:

#include "stdafx.h"
#include "stdafx.h"
#include <string>
#include <sstream>
#include <iostream>


using namespace std;

int strlen1( string str)
{
    int count=0;
    int i=0;
    while (str[i]!='\0')
        {
            count++;
            i++;
    }
    return count;
}

int _tmain(int argc, _TCHAR* argv[])
{
    string input;
    cin >> input;
    cout << strlen1(input) << endl;

    return 0;
}

thanks!

why `count++; i++;`? You only need one... but that's not the problem... — Federico klez Culloca, Feb 03 '11 at 19:34
Have you seen the std::string documentation? http://www.sgi.com/tech/stl/basic_string.html — Martin York, Feb 03 '11 at 19:40
@PigBen: That doesn't compile. You need: `int i ; for (i=0; i — TonyK, Feb 03 '11 at 19:44
This is wrong on so many levels, I wouldn't know where to start telling you. __Get a book!__ http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list — sbi, Feb 03 '11 at 20:52

Cheers and hth. - Alf · Accepted Answer · 2011-02-03T19:54:41.387

4

In short, for a non-const std::stringaccessing an element beyond the string length has Undefined Behavior (more precisely, when using the non-const operator[] for that).

C++98 §21.3.4/1, about std::string::operator[]:

If pos < size, returns data()[pos]. Otherwise, if pos == size(), the const version returns charT(). Otherwise the behavior is undefined.

And your string is non-const.

You can make it const, which will then invoke a special provision in the standard that guarantees zero result for element n, but that does not address the problem that your function is completely redundant for std::string.

Perhaps you mean to have argument type char const*.

That would make more sense, and then also the function would work.

Cheers & hth.,

edited Feb 03 '11 at 19:54

answered Feb 03 '11 at 19:40

Cheers and hth. - Alf

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the const made it work, but as you said i think that the right solution is to have argument type char* instead of string. unfortunately, when i replaced it in str and input variables it didnt work saying the variable 'input' is being used without being intialized. i have no idea how to make it work – Aviran Cohen Feb 03 '11 at 20:03
Just change the `str` variable to a `const char *`; leave `input` as a `string`. Then call it like `strlen1(input.c_str())`. – Daniel Gallagher Feb 03 '11 at 20:10
Never knew it. Apparently in C++0x `s[s.size()]` will always return a reference to a NUL character (and you must not modify the value). – UncleBens Feb 04 '11 at 00:05

score 3 · Answer 2 · answered Feb 03 '11 at 19:33

3

You seem to be under the misconception that std::string is the same as a C-style string (char const*). This is simply not true. There is no null terminator.

Here's your strlen function the only way it will work:

int strlen1(string str) { return static_cast<int>(str.size()); }

answered Feb 03 '11 at 19:33

Edward Strange

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Actually there is effectively a null-terminator, as soon as you make the string `const`. – Cheers and hth. - Alf Feb 03 '11 at 19:41
Not sure why you'd say constness matters here. Null is not considered a terminator for std::string. – Edward Strange Feb 03 '11 at 20:09
I did. It's right in that Null is returned, but you are in error here when you call it a terminator. – Edward Strange Feb 03 '11 at 21:38
@Noah Roberts: const matters because strlen1 should proudly announce to the world it won't destroy the string to get it's size... as does size() method prototype `size_t size() const;` – kriss Feb 03 '11 at 21:41
@kriss - sure, but it doesn't magically make 0 a terminator in std::string. – Edward Strange Feb 03 '11 at 21:48
@Noah: "in error when you call it a terminator", <-- technobabble. – Cheers and hth. - Alf Feb 03 '11 at 21:51
Calling it that doesn't make your statements accurate. – Edward Strange Feb 03 '11 at 22:01
@Noah: what's wrong? why can't you take a simple technical correction? – Cheers and hth. - Alf Feb 04 '11 at 12:20

score 1 · Answer 3 · answered Feb 03 '11 at 19:33

1

std::string has a nice size() method for this.

answered Feb 03 '11 at 19:33

Nikolai Fetissov

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are you sure you have to create a strlenb function that takes a std::string as an argument, not a `char*` ? – kriss Feb 03 '11 at 19:36
you right, i've made mistake, need it as char*. unfortunately, when i replaced it in str and input variables it didnt work saying the variable 'input' is being used without being intialized. i have no idea how to make it work – Aviran Cohen Feb 03 '11 at 20:04

score 0 · Answer 4 · answered Feb 03 '11 at 19:32

0

strlen is based on zero terminated strings (char*, or char arrays), C++ string are based on length + data, you can't access to data after the end of string (where a terminal zero would be). Beside that, getting the length is a standard property of C++ strings (size()), hence the standard way of writing strlen is just to call size.

answered Feb 03 '11 at 19:32

kriss

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Actually, for a *constant* `std::string` of length *n*, accessing element *n* (which is beyond the string length) is required to yield `'\0'`. It's a special case provision in the standard. So if the OP had just added a little `const` to the formal argument type the `strlen1` function would have been technically OK. – Cheers and hth. - Alf Feb 03 '11 at 19:35
what do you mean sero teminated strings? and how do you think i can create a strlen function by my own? – Aviran Cohen Feb 03 '11 at 19:37
@Aviran: I mean creating a strlen function based on std::string as input is probably one of dumber thing I ever heard of. If it's some homework you should consider looking for a better teacher. – kriss Feb 03 '11 at 21:21
@Alf P. Steinbach: technically ok, you say it, and there is many ways to do the same thing that would work without const. But I agree for a function like strlen, making parameter const looks logical, I wouldn't like a strlen that is allowed to destroyed the original string to compute the length. – kriss Feb 03 '11 at 21:35

Function that will act like an Strlen C++

4 Answers4