This program is not printing whole string

Question

In this program,I store integer values from 0 to 4 in a string and I want to print this string.

CODE:

int main()
{
 string str=" ";
 for(int i=0;i<5;i++)
   str[i]=i+'0';
 cout<<str;
 return 0;
}

But when i run this program,it just shows first element(means '0') of string.

Can anybody tell me the reason,Why this program does not print whole string in output?

Answer 1

The code doesn't work as expected because you exceeded the string greatest index which is 0 because the size of your string is 1 as the previous answer say.

This is a modification

int main()
{
    std::string str="     ";
    for(int i=0;i<5;i++)
        str[i]=i+'0';
    std::cout<<str;

    return 0;
}

Answer 2

Actually, when you are using for loop, it is accessing position in string changing it. Since when you have initialized string as string str = " " this means our string is of single whitespace. After which for loop access that string and look for the index str[i] and changes its value to i+'0'.

So, when we are writing code as

for(int i=0; i< 5; i++){
  str[i]=i+'0';
  cout<<str[i];
}

it seems, like for that block, str[i] value is i+'0', which is printed, but str doesn't update. Due to which when you do cout<<str; it again shows 0 as the answer.

To solve this issue we can use property of concatenation, refer the below code

 #include <iostream>
 using namespace std;

 int main()
 {
 string str;
 for(int i=0;i<5;i++)
    str+=i+'0';
 cout<<str;
 return 0;
 }

This gives the required result as 01234.

Hope this helps you.

Answer 3

You are exceeding array boundaries, because you initialized the string as containing only one space. If you give it more spaces, the code works:

string str = "     ";

Answer 4

when we are assigning

string str = " "; 

here " " is of type const char[2];size is 2: 1 lenght of character and 1 length for '\0' (Null Terminated) character. This gets copied into string str by calling operator= overloaded function.

When We are trying to access str[] with more than 0,1 index of subscript[] say 2 its doing memory violation and that is causing the exception "string subscript out of range",

if you try to access [0] and [1] it will work fine or instead of accessing particular location using array subscript, concatenation (+=) will help.

Answer 5

int main()
{
 string str=" ";        // str.size() == 1

 for(int i=0;i<5;i++)
   str[i]=i+'0';        // str.size() is still 1

 cout<<str;             // str.size() is still 1 
}

In str[i], if ...

• i>str.size() you are accessing the array out of bounds and get undefined behaviour.

• i==str.size() (where the null terminator is) and you change that value, you also get undefined behaviour.

But when i am printing this string with the help of for loop,this program prints whole string.e.g:

for(int i=0; i<5; i++) {
    cout<<str[i]<<" "; 
}

Why this loop print whole string, but not cout<<str?

You've made changes out of bounds and have undefined behaviour. It's possible that you can read the values back one-by-one (instead of getting a program crash) due to short string optimization - but the size() of str is still 1 since you haven't changed the size.

If your loop had gone further, outside the area used by short string optimization, like this:

for(int i=0; i<25; i++) {
    cout<<str[i]<<" "; 
}

it would be even more likely to result in a crash.