CODEWITHSUNDEEP
c++
Huppo
Huppo

Reputation: 971

confusing mixing datat type in c++

I am trying to figure out what is happening this code

int i = 10, j = 7, l;
unsigned int k;
double q = 3.56;
char c;

c = q * i * j;
l = c;
std::cout << l << "\n"; // result is -7
std::cout << c << "\n"; // result is ?

I couldn't understand why l result is '7', and for the c whatever i change the value i, j the result for c is always '?'. Could anyone point me a reason about that ? Thanks

Upvotes: 1

Views: 79

Answers (4)

Mark Byers
Mark Byers

Reputation: 838376

The result of 3.56 * 10 * 7 is 249.2.

If the type char on your system is a signed 8 bit value with a range from -128 to 127 then attempting to assign 249.2 will cause an overflow. If you assign to an int instead of a char it will be stored as 249.

The question mark is because -7 is not a printable character.

Upvotes: 6

Shawn Buckley
Shawn Buckley

Reputation: 516

3.56*7*10 is 249.200.....2. It becomes -7 when it is put into a char because a char is an 8 bit signed value. -7 comes from 249 (truncated) - 256.

Upvotes: 0

peacemaker
peacemaker

Reputation: 2591

You're writing the value 249.2 into a signed char which is out of the range, so it's converted to the ? character.

Upvotes: 0

Me.Name
Me.Name

Reputation: 12544

The default display method for a char is its ascii/unicode value. For values it cannot display, a ? is shown. For others the equivalent character would be shown. For example, a char value of 65 would show A

Upvotes: 1

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