length of char array using dynamic allocation

Question

Here is my code

#include<iostream>
#include<string>
using namespace std;
char* getString(char*& s)
{
    string str;
    getline(cin, str);
    int length = str.length();
    s = (char*)malloc(length);
    for (int i = 0; i < length; ++i)
        s[i] = str[i];
    return s;
}

void main()
{
    char* s;
    getString(s);
    cout << s << endl;
}

In my getString() function, I create a string str and pass the input string to it. Then I calculate for the length of str and then allocate this number of bytes for the char array s. But the problem is, when i call for strlen(s), the return value is not equal to the length of string str, so when i print out s to the console, it's not the string i want. Anyone can tell me why and explain more about it??

This is C++ code, not C code, and should be tagged with the C++ tag, not the C tag. — Eric Postpischil, Aug 06 '20 at 02:31
This is just for practice, yes? In ordinary code there will not ordinarily be a reason to do this - just pass the actually `std::string` around. — Karl Knechtel, Aug 06 '20 at 02:39
Does this answer your question? [How to convert a std::string to const char\* or char\*?](https://stackoverflow.com/questions/347949/how-to-convert-a-stdstring-to-const-char-or-char) — Karl Knechtel, Aug 06 '20 at 02:40
@KarlKnechtel yes, my instructor asked me to use the dynamic allocation the solve the task. — babylearnmaths, Aug 06 '20 at 02:45
If you have to do things a specific way because of an assignment specification, you should explain this up front. — Karl Knechtel, Aug 06 '20 at 03:11

score 2 · Answer 1 · edited Aug 06 '20 at 02:37

2

You need to allocate one more character than str.length() to allow for the null character that terminates the string, and you need to set that character to the null character.

edited Aug 06 '20 at 02:37

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answered Aug 06 '20 at 02:32

Eric Postpischil

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Also, if this answer worked for you, please accept it – kesarling He-Him Aug 06 '20 at 02:38

score 1 · Accepted Answer · answered Aug 06 '20 at 02:43

When you have an array of characters the very last byte is called the null terminator, which is \0 in this case. Its what tells your program where the end of the characters in an array are.

The str.length() function returns how many bytes are in a string. A null character does not count as a byte. So when str.length() goes to count how many characters are in your array, it counts until it reaches \0.

When you need to create a new array of the same 'size' as the string, you must make it one byte longer in order to accommodate the null terminator.

So your code should be:

s = (char*)malloc(length + 1);

length of char array using dynamic allocation

2 Answers2