Difference in type conversion between C and C++?

Question

The last example in this tutorial regarding Implicit type conversion states that std::cout << 5u - 10; will produce 4294967291 rather than -5 due to type conversion. I tried this in both C and C++. The result in C++ was as promised, however when using C ( printf("%d\\n", 5u - 10); ) the result was -5 . What happened?

Answer 1

In the C example there is no type conversion whatsoever. C just evaluates the expression 5u - 10 and pushes the result onto the stack. Then printf sees a type character and interprets the value on the stack as such when printing it. The type character is d ( %d ) meaning "decimal integer" and hence the position on the stack is retrieved as an int and is printed as (signed) decimal.

Would the type character be for example ld ( %ld ), the position on the stack would be retrieved as a long, even if only an int was pushed, and that would be printed as a (signed) decimal number. Again, there is no type conversion whatsoever (there will just be a nonsense number printed).

Answer 2

A little more thought brought upon the realization that the problem was with the printf rather than the conversion itself. Notice that what was written was printf("%d") . This performed yet another conversion back to signed int , which is why I saw the result of -5 .

When testing with printf("%u") , the promised result ( 4294967291 ) was shown.

To cap it off, printf("%X") resulted in FFFFFFFB which means both those values, depending on a signed or unsigned interpretation.