Declaring a variable with two types: “int char”

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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?


int


char





Which edition and printing of the book do you have? Have you looked for an errata of the book?
– Some programmer dude
Jul 9 at 6:57





@Someprogrammerdude thanks for the reminder. I should definitely check the errata first.
– Thor
Jul 9 at 6:58





So what version are you reading? I'm sure Bjarne would like to know about this mistake.
– StoryTeller
Jul 9 at 7:03





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
Jul 9 at 7:06





The float char is another useful type, especially in swimming pools. Some come with a holder for a beer.
– Yakk - Adam Nevraumont
Jul 9 at 16:01




float char




4 Answers
4



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.


//error





What was probably the intended code example in the book?
– Hamsterrific
Jul 9 at 21:57





@Hamsterrific char b1 {1000}; (because that would provoke the error mentioned in the comment). I guess Bjarne's typing fingers were tired that day.
– Paul Sanders
Jul 9 at 22:53




char b1 {1000};





@PaulSanders Tired? He typed an extra int there! :-)
– Leo Heinsaar
Jul 10 at 7:11


int





@ LeoHeinsaar Lol. OK, too much coffee then :) Or maybe he has a stutter :)
– Paul Sanders
Jul 10 at 7:47





Yeah: fingers tired. It looks like a cut-n-paste error from the example that preceded it. In that prior example, he gives two lines: int a {1000}; // OK [n] char b {a} // error: int -> char might narrow [n] and he seems to have cut-n-pasted that example for the next one and missed deleting the "int" part: int char b1 {1000}; // error: narrowing (assuming 8-bit chars) [n] char b2 {48}; // OK [n]
– L. Scott Johnson
Jul 13 at 18:01





The book is wrong.



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


int char b1{1000};



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


b1



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.



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


union


std::variant


#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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