Declaring a variable with two types: "int char"

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?

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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.

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最佳答案

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.

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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.

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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
}

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