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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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How can a variable have 2 data types? It cannot. An object has one type. A single type can represent values of different types (unions, std::variant ).
A variable can be declared as double by adding the double keyword as a prefix to it. You majorly used this data type where the decimal digits are 14 or 15 digits.
If a value is to be assigned at the time of declaration, you can use this syntax: char variable-name = 'value'; The variable-name is the name of the char variable. The value is the value to be assigned to the char variable.
If we direct assign char variable to int, it will return the ASCII value of a given character. If the char variable contains an int value, we can get the int value by calling Character. getNumericValue(char) method. Alternatively, we can use String.
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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