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What's the difference between the types of bar1 and bar2?
int foo = 10;
auto bar1 = &foo;
auto *bar2 = &foo;
If both bar1 and bar2 are int*, does it makes sense to write the pointer declarator (*) in the bar2 declaration?
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C++11 introduces the keyword auto as a new type specifier. auto acts as a placeholder for a type to be deduced from the initializer expression of a variable. With auto type deduction enabled, you no longer need to specify a type while declaring a variable.
The auto keyword directs the compiler to use the initialization expression of a declared variable, or lambda expression parameter, to deduce its type.
auto_ptr is a class template that was available in previous versions of the C++ standard library (declared in the <memory> header file), which provides some basic RAII features for C++ raw pointers. It has been replaced by the unique_ptr class.
Note that the asterisk ( * ) used when declaring a pointer only means that it is a pointer (it is part of its type compound specifier), and should not be confused with the dereference operator seen a bit earlier, but which is also written with an asterisk ( * ).
The declarations are exactly equivalent. auto works (almost) the same as template type deduction. Putting the star explicitly makes the code a bit easier to read, and makes the programmer aware that bar2 is a pointer.
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Using auto * "documents intention". And auto *p = expr; can be deduced correctly only if expr returns pointer. Example:
int f();
auto q = f(); // OK
auto *p = f(); // error: unable to deduce 'auto*' from 'f()'
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