Declaring a variable with two types: "int char"

81
votes

I'm a C++ beginner, and I'm reading Bjarne Stroustrup's Programming: Principles and Practice Using C++.

In the section on 3.9.2 Unsafe conversions, the author mentioned

When the initializer is an integer literal, the compiler can check the actual value and accept values that do not imply narrowing:

int char b1 {1000};     // error: narrowing (assuming 8-bit chars)

I'm puzzled by this declaration. It uses two types (int and char). I have never seen such declaration in Java and Swift before (the two languages I'm relatively familiar with). Is this a typo or a valid C++ syntax?

4
c++type-conversioninitializationuniform-initializationnarrowing
Which edition and printing of the book do you have? Have you looked for an errata of the book? - Some programmer dude
So what version are you reading? I'm sure Bjarne would like to know about this mistake. - StoryTeller - Unslander Monica
3.9.2 Unsafe conversions By unsafe conversion we mean that a value can be implicitly turned into a value of another type that does not equal the original value. e.g: int i = 20000; char c = i; Such conversions are called 'narrowing' conversions. double to int, char or bool int to char or bool char to bool - Marichyasana
The float char is another useful type, especially in swimming pools. Some come with a holder for a beer. - Yakk - Adam Nevraumont
Is this a typo or a valid C++ syntax? Try it (OK, OK, no it isn't valid). - Paul Sanders

4 Answers

95
votes

It's a mistake in the book. That is not a valid C++ declaration, even without the supposed narrowing conversion.

It isn't mentioned in any of the erratas on Bjarne Stroustrup's page(4th printing and earlier), though, which is odd. It's a clear enough mistake. I imagine since it's commented with //error few people notice the mistake in the declaration itself.

24
votes

The book is wrong.

The token sequence int char b1{1000}; is not semantically valid C++.

You are trying to declare b1 with more than one type, which makes no sense.

10
votes

It is wrong. In C/C++ the multi-type declarations can be achieved via the use of unions. Eg:

union {
    int i;
    char c;
} var;

var.i = 42;
/* OR */
var.c = ‘c’;

The storage is the same, so .c and .i are just per-type handles to the same value.

6
votes

This is wrong in C/C++ syntax. In addition to unions (see @Alex answer), there is a C++ way to store only one of available types called std::variant (type-safe union):

#include <variant>
#include <string>

int main()
{
    std::variant<int, float> v, w;
    v = 12; // v contains int
    int i = std::get<int>(v);
    w = std::get<int>(v);
    w = std::get<0>(v); // same effect as the previous line
    w = v; // same effect as the previous line

//  std::get<double>(v); // error: no double in [int, float]
//  std::get<3>(v);      // error: valid index values are 0 and 1

    try {
      std::get<float>(w); // w contains int, not float: will throw
    }
    catch (std::bad_variant_access&) {}

    std::variant<std::string> v("abc"); // converting constructors work when unambiguous
    v = "def"; // converting assignment also works when unambiguous
}