Type conversion char to int to use as an index of a vector C++

Question

In my program I have a string of caracters that are all digits and I want to use them as indexes to get values from a vector. But the output of this code is a 15 digit number, so not 1 which is vect[0] i.e. what I thought it would output.

char c = '0';
vector<int> vect = {1};
int i = (int)c;
cout<<vect[i];

I have seen that converting from a caracter to an integer is done with

int i=(int)c;

So i really don't get how that is not an appropriate type for an index.

`int i = (int)c` does not results in `i=0`, but int `i=48`. 48 is the position of char '0' in the [ASCII table](http://www.asciitable.com/). You want `int i = (int)c - 48` — Ripi2, Oct 26 '18 at 17:29

Matthieu Brucher · Answer 1 · 2018-10-26T17:49:52.803

1

First, it is bad practice to use C-style casts (int i = (int) c;), use a static cast instead (int i = static_cast<int>(c);). But indeed, as other mentioned, char are promoted to int automatically.

Then character '0' is not the same as the index 0. Have a look:

int i = c;
std::cout << i << std:endl;

If you want to use the character '0' as index 0, use:

int i = c - '0';

edited Oct 26 '18 at 17:49

answered Oct 26 '18 at 17:29

Matthieu Brucher

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But since c is '0' isn't that always going to be 0? – John Cataldo Oct 26 '18 at 17:32
There's no need to cast at all. All `char`'s get promoted to `int`'s during arithmetic. `int i = c - '0';` is all that is needed. – NathanOliver Oct 26 '18 at 17:32
@NathanOliver Indeed. – Matthieu Brucher Oct 26 '18 at 17:50
@JohnCataldo depends on what `c` is going to be later? Otherwise, what's the point of the code? – Matthieu Brucher Oct 26 '18 at 17:50

Type conversion char to int to use as an index of a vector C++

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