Looping through string of integers gives me completely different numbers? - c++

Question

I'm a beginner to C++ so forgive me if I'm making a stupid mistake here.
I want to loop through a string of integers in the following code:
#include <string>
using namespace std;
int main() {
string str = "12345";
for (int i : str) {
cout << i << endl;
}
return 0;
}
But I receive the output:
49
50
51
52
53
I know that I get normal output if I use char instead of int, but why do I receive an output of integers 48 more than they should be?

Answer 1

When you loop through a string you get elements of type char. If you convert a char to an int you get the ASCII value of the char, which is what happens when you do:
string str = "12345";
for (int i : str) { // each char is explicitly converted to int
cout << i << endl; // prints the ascii value
}
The ASCII value of '0' is 48, and '1' is 49, etc, which explains the output you get.

Answer 2

Just what #cigien said, You just need to change it from int to char i.e
string str = "12345";
for (char i : str) {
cout << i << endl;
}
Or one solution for all auto keyword
string str = "12345";
for (auto i : str) {
cout << i << endl;
}

Answer 3

The first thing you need to know is that a string is an array/sequence of chars.
You can think of a char as a single character.
But the way it is encoded is as a number.
For example, the char 'a' is encoded (in ASCII) as the number 97.
Now your for loop says int i: str.
You're telling it to look for integers in the string.
But a string is an array/sequence of chars, not of integers.
So the loop takes each char,
and instead of looking at what the character itself is,
it gives you the integer encoding value of the char,
the ASCII value.
Now the numbers are encoded with the char '0' having the lowest encoding value,
'1' having the next value,
'2', having the next,
and so on through digit '9'.
I can never remember what the actual ASCII value for '0' is . . . .
But because the digit chars are encoded consecutively in this way,
you can convert any digit char to its int value by subtracting the underlying integer encoding value of '0'.
#include <string>
using namespace std;
int main() {
string str = "12345";
for (char c: str) {
cout << (c - '0') << endl; // gives you the actual int value, but only works if the char is actually a digit
}
return 0;
}

Answer 4

for (int i : str) { is infact syntactic sugar for
for (auto iterator = str.begin(); iterator != str.end(); iterator++) {
int i = (int) *iterator;
But the *-operator from string::iterator is infact an overload which returns the current char. It will as such be casted to an int. What you then see is this number. It is the integer value of the byte. Not necessarily ASCII. It could be ANSI too.