char ch=”a”; std::cout << ch << ‘\n’; std::cout << +ch << ‘\n’; The first insertion writes the character a to cout. The second insertion writes the numeric value of ch to cout. But that’s a bit obscure; it relies on the compiler applying integral promotions for the + operator.
In both cases, the last rofl is the variable name. A variable name is in scope as soon as it appears; and for the remainder of the current scope, that identifier in an ordinary context(*) always means the variable name. The sizeof operator does not introduce any special cases for name lookup. In fact, there … Read more
Yes, 6.2.5c9 is exactly the paragraph that you looked for.
The unary plus operator performs an automatic conversion to int when the type of its operand is byte, char, or short. This is called unary numeric promotion, and it enables you to do things like the following: char c=”c”; int i = +c; Granted, it’s of limited use. But it does have a purpose. See … Read more
Built-in operator+ could take pointer type as its operand, so passing the array s to it causes array-to-pointer conversion and then the pointer int* is returned. That means you might use +s individually to get the pointer. (For this case it’s superfluous; without operator+ it’ll also decay to pointer and then assigned to p.) (emphasis … Read more
Use the format specifier %p: printf(“variable A is at address: %p\n”, (void*)&A); The standard requires that the argument is of type void* for %p specifier. Since, printf is a variadic function, there’s no implicit conversion to void * from T * which would happen implicitly for any non-variadic functions in C. Hence, the cast is … Read more
You can use it as a sort of assertion that an expression has arithmetic type: #define CHECK_ARITHMETIC(x) (+(x)) This will generate a compile-time error if x evaluates to (say) a pointer. That is about the only practical use I can think of.
Actually, unary plus does do something – even in C. It performs the usual arithmetic conversions on the operand and returns a new value, which can be an integer of greater width. If the original value was an unsigned integer of lesser width than int, it will be changed to a signed value as well. … Read more
The output operator << when being passed a char (signed or unsigned) will write it as a character. Those function will return values of type unsigned char. And as noted above that will print the characters those values represent in the current encoding, not their integer values. The + operator converts the unsigned char returned … Read more
Same as in other languages: ++x (pre-increment) means “increment the variable; the value of the expression is the final value” x++ (post-increment) means “remember the original value, then increment the variable; the value of the expression is the original value” Now when used as a standalone statement, they mean the same thing: x++; ++x; The … Read more