Copying a string char by char in another string

Question

I have a string type and I am trying to copy this into another string type char by char but when I try to display the output of new copied it is coming as blank. Where as when I try to output char by char for that copied string length its alright. Please see this small code and output for better understanding.

#include <iostream>
#include <string>

using namespace std;

int main()
{
    string name = "Tom";
    string temp;
    int i = 0;
    for(; i < name.length(); ++i)
    {
        temp[i] = name[i];
    }
    cout<<name<<endl;     //gives the output Tom
    cout<<temp<<endl;     //gives blank
    for(int i = 0; i < name.length(); ++i)
    {
        cout<<temp[i];    //gives output char by char
    }
}

Output :

Tom

Tom

I know there are other possible ways to achieve what I am trying to do but just out of thought tried this and it didn't work. So looking for some explanation.

temp has zero length, so you cannot use indexing to address any of its characters. Use temp.push_back(name[i]); to make temp grow. — Christopher Oicles
– Christopher Oicles, Commented May 16, 2017 at 0:32
Usiing at() shows your code will not work due to you going out of bounds of the string — PaulMcKenzie
– PaulMcKenzie, Commented May 16, 2017 at 0:39

songyuanyao · Accepted Answer · 2017-05-16 00:59:43Z

2

Note that temp is empty, its length is 0. Then temp[i] = name[i]; leads to undifined behavior; anything is possible but nothing is guaranteed.

You can give it an initial length like:

string name = "Tom";
string temp(name.length(), '\0');

edited May 16, 2017 at 0:59

answered May 16, 2017 at 0:35

songyuanyao

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2 Comments

rakesh sinha Over a year ago

I get that, still not clear as why in for loop it was giving right answer.

songyuanyao Over a year ago

@rakeshsinha Undefined behavior means anything is possible; but nothing is guaranteed.

Native Function · Accepted Answer · 2017-05-16 00:35:27Z

0

You have to allocate data for the temp string to change any data in it.

string temp(name.length(),'\0');

answered May 16, 2017 at 0:35

Native Function

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Comments

user2672107 · Accepted Answer · 2017-05-16 11:22:13Z

0

for(; i < name.length(); ++i) temp.push_back(name[i]);

Don't assign.

answered May 16, 2017 at 11:22

user2672107

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