How to get int from string (string using like a array of char)

Question

this little program describes my problem which I have in bigger project:

int main()
{
    string str1="11111111";
    string str2="22222222";

    char a=str1[2];
    char *a1=&a;
    int var1= atoi(a1);

    char b=str2[2];
    char *b1=&b;
    int var2= atoi(b1);

    cout<<var1<<endl;
    cout<<var2;


    return 0;
}

Why I'm getting

1
21

istead of

1
2

?

Is there any way to fix this? - thanks for the help I trying to figureout for two hours

Strings need to end with the special '\0' character, I believe. — nico, Apr 25 '17 at 16:29
If all the characters in the string are numbers, [you can just subtract `'0'` (character literal, not integer literal)](http://ideone.com/BonUO1), assuming a character set where all numbers are listed sequentially. — Justin Time - Reinstate Monica, Apr 25 '17 at 16:42
Also, end the program with another endl after var2 is printed to avoid runtime problems. — Jish, Apr 25 '17 at 23:14

score 2 · Accepted Answer · answered Apr 25 '17 at 16:33

2

You get both results by mistake (even though your first result happens to match your expectation).

The problem is that both a1 and b1 point to a single character, while atoi expects a null-terminated string.

You can fix this problem by constructing character arrays instead of copying individual characters:

char a1[2] = { str1[2] };   // C++ compiler supplies null terminator, you supply the size
int var1= atoi(a1);
char b1[] = { str2[2], 0 }; // You supply null terminator, C++ compiler computes the size
int var1= atoi(b1);

answered Apr 25 '17 at 16:33

Sergey Kalinichenko

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Thanks so much I used your advice to get numbers in loop: https://pastebin.com/wMqV7vZm – Adrian Apr 26 '17 at 16:42
Sorry previous code wasn't good enough - this work perfectly https://ideone.com/u5YGLc – Adrian Apr 26 '17 at 16:54

score 1 · Answer 2 · answered Apr 25 '17 at 16:36

Use std::stoi() and std::string::substr() especially if you have std::string already:

std::string str1="11111111";
std::string str2="22222222";

int var1= std::stoi( str1.substr( 2, 1 ) ); // third symbol, 1 symbol long

int var2= std::stoi( str2.substr( 2, 1 ) );

live example

score 0 · Answer 3 · edited May 23 '17 at 11:46

0

atoi expects a pointer to a null-terminated string of char. You pass a pointer to a char. What happens is undefined. You better use std::stoi instead of atoi since atoi has some fallacies: What is the difference between std::atoi() and std::stoi?

edited May 23 '17 at 11:46

Community

1
1

answered Apr 25 '17 at 16:30

The Techel

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8
13

score 0 · Answer 4 · answered Apr 25 '17 at 16:32

0

atoi wants a pointer to the first element of a zero-terminated sequence of characters.
You're passing it a pointer to a single character, leaving undefined behaviour in your wake.

To get the integer value of one of the digits, take its distance from '0':

int var = str1[2] - '0';

answered Apr 25 '17 at 16:32

molbdnilo

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Unfortunatelly my teacher doesnt allow us to use this methon :/ I have to use atoi :( – Adrian Apr 26 '17 at 16:44
@AdrianWąt That kind of information belongs in the question. – molbdnilo Apr 26 '17 at 16:45

How to get int from string (string using like a array of char)

4 Answers4