Cannot convert string to const char

Question

I have this function and the compiler yells at me saying "Cannot convert string to const char".

void 
DramaticLetters(string s, short TimeLength)
{
    for(int i = 0, sLen = strlen(s); i < sLen; i++){
        cout << s[i];
        Sleep(TimeLength);
    }
}

There's something wrong with the strlen, I think

Just use `s.length()` or `s.size()`. Strings don't implicitly convert to character arrays. If you really need one (which you don't for length), use `s.c_str()`. — chris, Aug 15 '12 at 14:09
http://stackoverflow.com/questions/347949/convert-stdstring-to-const-char-or-char?rq=1 — Matt K, Aug 15 '12 at 14:10
Just a general hint - if something goes wrong with standard functions, 99.99% of the time it's your fault. — Cubic, Aug 15 '12 at 14:12
Your un-use of the member functions makes me want to guide you to this [reference page](http://en.cppreference.com/w/cpp/string/basic_string). Look through and get a feel of what's available. — chris, Aug 15 '12 at 14:12
@Cubic: Exactly. I've had experiences where something *had* to have been an standard lib error, but that was an error in my code. Anything you think is a bug is most likely implementation defined behavior. Just read the docs. — Linuxios, Aug 15 '12 at 14:17

Linuxios · Accepted Answer · 2012-08-17T14:50:47.067

strlen() is for C const char* strings. To get the length of a string s, use s.size() or s.length(). If you want to get a C string from a string, use s.c_str().

Although C++ makes it seem that const char* and string are interchangeable, that only goes one way when converting const char* to string.

There is no reason why you would want to use strlen either. strlen is most likely defined with a loop, which will never be as efficiant as size(), which is most likley just a getter for a length property of the string class. Only convert string to C strings when calling C functions for which there is not a C++ alternative.

score 1 · Answer 2 · answered Aug 15 '12 at 14:12

1

You should not mix C and C++ string functions. Instead of strlen() (a C-style function), use string::size():

for(int i = 0, sLen = s.size(); i < sLen; i++) {
    cout << s[i];
    Sleep(TimeLength);
}

Here you have a reference with all methods from class string.

answered Aug 15 '12 at 14:12

betabandido

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So not all C functions can be used in C++? – skittles sour Aug 16 '12 at 01:34
@user1575615 C++ "reimplements" many of the C features (e.g., using class `std::string` instead of just a char array and `strlen`, `strcpy`, etc.). You can still use the C-style features, though. Actually you can make a C++ program that looks like a C one, but you should not mix C and C++ solutions, such as `std::string` (from C++) with `strlen`, `strcmp`, etc. (from C). – betabandido Aug 16 '12 at 08:06

score 1 · Answer 3 · answered Aug 15 '12 at 14:12

As chris said in his comment, strlen is used for a const char *, whereas you're passing it a string. Instead of

for(int i = 0, sLen = strlen(s); i < sLen; i++){
    cout << s[i];
    Sleep(TimeLength);
}

Use this:

for(int i = 0, sLen = s.length(); i < sLen; i++){
    cout << s[i];
    Sleep(TimeLength);
}

score 0 · Answer 4 · answered Aug 15 '12 at 14:12

0

strlen operates on char const *. If you really wanted to, you could do strlen(s.c_str()), but std::string has a lot of functionality, including a length() method, which returns the number of characters in the string

answered Aug 15 '12 at 14:12

Tom Tanner

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score 0 · Answer 5 · answered Aug 15 '12 at 14:12

Few remarks:

You do not modify the string within the function so better pass a const reference to it(const string& s)
The string itself defines methods for getting its length - s.size() and s.length() both will work. Additional benefit is both this methods are constant with respect to complexity as opposed to the linear complexity of strlen
If you REALLY want to use strlen use s.c_str(): strlen(s.c_str())

Cannot convert string to const char

5 Answers5