unsigned char to int in C++

Question

I have a variable unsigned char that contains a value, 40 for example. I want a int variable to get that value. What's the simplest and most efficient way to do that? Thank you very much.

Solution 1

unsigned char c = 40;
int i = c;

Presumably there must be more to your question than that...

Solution 2

Try one of the followings, it works for me. If you need more specific cast, you can check Boost's lexical_cast and reinterpret_cast.

unsigned char c = 40;
int i = static_cast<int>(c);
std::cout << i << std::endl;

or:

unsigned char c = 40;
int i = (int)(c);
std::cout << i << std::endl;

Solution 3

Google is a useful tool usually, but the answer is incredibly simple:

unsigned char a = 'A'
int b = a

Solution 4

Depends on what you want to do:

to read the value as an ascii code, you can write

char a = 'a';
int ia = (int)a; 
/* note that the int cast is not necessary -- int ia = a would suffice */

to convert the character '0' -> 0, '1' -> 1, etc, you can write

char a = '4';
int ia = a - '0';
/* check here if ia is bounded by 0 and 9 */

Solution 5

Actually, this is an implicit cast. That means that your value is automatically casted as it doesn't overflow or underflow.

This is an example:

unsigned char a = 'A';
doSomething(a); // Implicit cast

double b = 3.14;
doSomething((int)b); // Explicit cast neccesary!

void doSomething(int x)
{
...
}

View more solutions

Comments

[user1346664]

if I write printf("c: %x", c) I get 32. Then i write int i=c and when I print "i" I get 50!

[Oliver Charlesworth]

@user1346664: %x prints in hexadecimal. 32 hex is equal to 50 decimal. Use %d instead.

[chris]

Why do you need an explicit cast for double->int? Won't it just truncate implicitly?

[bytecode77]

Yes it does, but you'll get a compiler warning, so it is recommended to do that.

[chris]

Odd, I don't get any warnings on the highest level of GCC.

[Bo Persson]

The value of a (and of i) will be 40, as that is the value you just assigned.

[Bob Jarvis - Слава Україні]

If you ignore the compiler warning you'll discover that i gets the address where the string "40" is stored, which is of course what 'a' has in it. Ah - brings back to good ol' days before that wimpy ANSI standard thing - back when C was weakly typed, programmers had stronger minds, and Coca-Cola was made with real sugar! Bah! Kids these days..! Bah!!

[Aamir]

The question was about unsigned char. Why are you using a char*?

[jaques-sam]

Prefer C++ constructor style int(c) over C-cast style: (int)(c)