Why am I unable to pass an iterator to string as an argument to the function strcat()?

Question

I've been constantly getting this error for the following block of program, isn't an iterator also a pointer when using such string data types?

 error: cannot convert 'std::basic_string<char>::iterator 
 {aka __gnu_cxx::__normal_iterator<char*, std::basic_string<char> >}'
 to 'const char*' for argument '2' to 'char* strcat(char*, const char*)'
 strcat(temp, str1.begin());
                        ^





            int areRotations(string str1, string str2)
            {
              int size1   = str1.length();
              int size2   = str2.length();
              char* temp= new char[size1+size2+1];
              void *ptr;
              if (size1 != size2)
              return 0;


              strcat(temp, str1.begin());
              strcat(temp, str1.begin());

              /* Now check if str2 is a substring of temp */
               ptr = strstr(temp, str2);


               if (ptr != NULL)
               return 1;
               else
               return 0;
            }

Why do you use `strcat` with `std::string`? Or, rather, why isn't `temp` a `std::string`? — M Oehm, Mar 21 '16 at 09:25
This function boils down to a single line: `return str1.size() == str2.size() && (str1+str1).find(str2) != string::npos;` — paddy, Mar 21 '16 at 09:33
Iterators *behave* very much like pointers - but they don't (necessarily) have the same type as pointers. Your code has a bug given `areRotations( "abcdefghijklmnopqrstuvw", "1")` - you will overflow temp. Also you have a memory leak. You want `bool areRotations(const std::string& str1, const std::string& str2) { return (str1+str1).find(str2) != std::string::npos;}` — Martin Bonner supports Monica, Mar 21 '16 at 09:35
@MartinBonner Hey I've allotted temp variable a **space** of char* temp= new char[size1+size2+1]; Isn't that sufficient? — Yash Kant, Mar 21 '16 at 09:42
Nothing like receiving advice and then refusing to follow it. — paddy, Mar 21 '16 at 09:43
@paddy Hey I've solved it!! by passing the pointer to string as an argument....thanks :) — Yash Kant, Mar 21 '16 at 09:46
All other proposed solutions are better than what you just described. — paddy, Mar 21 '16 at 09:49
@paddy yes I get it exactly, but I was pondering over the properties of strcat() function....thanks again :V — Yash Kant, Mar 21 '16 at 09:51
Then you should probably accept the answer that mentions using `c_str` to obtain a pointer to the buffer, amongst other explanations. Don't forget to free your temporary memory with the correct array-delete operator on all control paths, or use `std::unique_ptr`. — paddy, Mar 21 '16 at 09:54
@YashKant: No. In my example, you would allows 23+1+1 characters, and then copy 23+23+1 characters in. That won't fit. *pause* Ah! I missed the `if (size1!=size2) return 0;`. In that case the allocation is confusing but safe (better would be `2*size1+1`). — Martin Bonner supports Monica, Mar 21 '16 at 09:55
@YashKant I did not down-vote. My comment was not ill-intended. I was simply wondering why you chose to use strcat with std::strings... — LiMuBei, Mar 21 '16 at 14:07

score 4 · Accepted Answer · answered Mar 21 '16 at 09:27

You ask,

” isn't an iterator also a pointer when using such string data types

No, not necessarily.

For std::string it can be a pointer because std::string has a guaranteed contiguous buffer, so incrementing a pointer moves it forward in the buffer. However, I don't know of any extant C++ standard library implementation where it is a pointer.

A simple way to get a pointer to the buffer, that works also for an empty string, is to use the data member function. When you know that the string is non-empty you can also use &s[0] for this. A simple way to get a pointer to non-modifiable zero-terminated string, is to use the c_str member function.

Instead of strcat, with std::string you can just use + or +=.

score 3 · Answer 2 · edited May 23 '17 at 12:00

Pointer are iterators. the other way around is not always true. You can find here more information: How are iterators and pointers related?

However you are mixing up quite some things... C, C++. A simple C++ solution to your problem which simply uses std::string would be:

#include <iostream>

using namespace std;
bool areRotations(const string& str1, const string& str2)
{
    const string temp = str1 + str1;
    return ((temp.find(str2)!=std::string::npos) && (str1.size()==str2.size()));
}

int main()
{
    const std::string str1 = "hello";
    const std::string str2 = "lohel";
    cout << areRotations(str1,str2) << endl;
}

The original had a test that str1 and str2 were the same length. You might want to add that. — Martin Bonner supports Monica, Mar 21 '16 at 09:57

Why am I unable to pass an iterator to string as an argument to the function strcat()?

2 Answers2