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
– Eric Postpischil, Commented Aug 6, 2020 at 2:31
If you have to do things a specific way because of an assignment specification, you should explain this up front. — Karl Knechtel
– Karl Knechtel, Commented Aug 6, 2020 at 3:11

bitmask · Accepted Answer · 2020-08-06 02:37:12Z

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 6, 2020 at 2:37

bitmask

35.2k17 gold badges107 silver badges174 bronze badges

answered Aug 6, 2020 at 2:32

Eric Postpischil

233k15 gold badges199 silver badges379 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

kesarling Over a year ago

Also, if this answer worked for you, please accept it

Ovcheric · Accepted Answer · 2020-08-06 02:43:50Z

1

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);

answered Aug 6, 2020 at 2:43

Ovcheric

1691 silver badge9 bronze badges

Collectives™ on Stack Overflow

length of char array using dynamic allocation

2 Answers 2

1 Comment

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related