-2 int() is the constructor of class int. It will initialise your variable a to the default value of an integer, i.e. 0. Even if you don't call the constructor explicitly, the default constructor, i.e. int() , is implicitly called to initialise the variable. Otherwise there will be a garbage value in the variable.
The internal representation of int and unsigned int is the same. Therefore, when you pass the same format string to printf it will be printed as the same. However, there are differences when you compare them. Consider: This can be also a caveat, because when comparing signed and unsigned integer one of them will be implicitly casted to match ...
64 I prefer int* i because i has the type "pointer to an int", and I feel this makes it uniform with the type system. Of course, the well-known behavior comes in, when trying to define multiple pointers on one line (namely, the asterisk need to be put before each variable name to declare a pointer), but I simply don't declare pointers this way.
int a = 5; int& b = a; b = 7; cout << a; prints out 7, and replacing int& b with int &b also prints out 7. In fact so does int&b and int & b. I tested this kind of behavior with a simple class as well. In general, does it ever matter whether the ampersand is placed relative to the type and identifier? Thanks.
int is a primitive data type while Integer is a Reference or Wrapper Type (Class) in Java. after java 1.5 which introduce the concept of autoboxing and unboxing you can initialize both int or Integer like this.
│int││int││int││int││int││int││ └───┘└───┘└───┘└───┘└───┘└───┘└┄ Of course, there's no reason they can't all point at the same int, or whatever. You may want to use an array of pointers if you want many pointers that you can easily iterate over.
In C#, int and Int32 appear to be the same thing, but I've read a number of times that int is preferred over Int32 with no reason given. Are the two really the same? Is there a reason where one sho...
7 c is a double variable, but the value being assigned to it is an int value because it results from the division of two int s, which gives you "integer division" (dropping the remainder). So what happens in the line c=a/b is a/b is evaluated, creating a temporary of type int the value of the temporary is assigned to c after conversion to type ...
29 int and unsigned int are two distinct integer types. (int can also be referred to as signed int, or just signed; unsigned int can also be referred to as unsigned.) As the names imply, int is a signed integer type, and unsigned int is an unsigned integer type.
