Cast from int to char

Question

I'd like to know how this cast works :

int value = 100;
auto f = [&value] (int x, int y) -> char { return x + y + value; };
printf("%d\n", f(10, 20));
value = 200;
printf("%d\n", f(10, 20));
return 0;

It prints -126 and -30 but i don't understand how is it possible to print a negative integer.

Take a peek at my answer to a very related question https://stackoverflow.com/a/56424762/276193 — learnvst, Jun 19 '19 at 20:06
It's implementation-defined whether `char` is signed or unsigned. If it is signed, overflow has undefined behaviour. — molbdnilo, Jun 19 '19 at 20:14
Change `char` to `unsigned char` if you want to represent 0 .. 255. — stark, Jun 19 '19 at 20:30
There are no casts in this code. The question seems to be about a **conversion**. A **cast** is something you write in your source code to tell the compiler to do a conversion l — Pete Becker, Jun 20 '19 at 03:17

score 0 · Answer 1 · answered Jun 19 '19 at 19:53

0

The numerical range for char is -128..127. 10 + 20 + 100 is 130 and outside of that range, so it wraps around when expressed as char.

To fix you need to use a data type that can contain values large enough:

auto f = [&value] (int x, int y) -> int { return x + y + value; };

Or use values < 127.

answered Jun 19 '19 at 19:53

tadman

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it's an example from my OOP course, not some code written by me. Can you please explain the calculation such that the example lead to -126 and -30? I didn't really understand how it "wraps" around – Jun 19 '19 at 20:06
1

Signed integer numbers in modern machines are stored in [Two's Complement](https://en.wikipedia.org/wiki/Two's_complement). An 8-bit value (`char`) can either be unsigned (`0`..`255`) or signed with Two's Complement (`-128`..`127`). No other values can be represented. In the case of signed, `char x = 127 + 1` assigns `-128` to `x`. In your case, 130 is precisely 3 over the limit, so it's the same as `127 + 3` where that's `-126` wrapped around. – tadman Jun 19 '19 at 20:13

Cast from int to char

1 Answers1