A C++ question, I know int* foo (void) foo will return a pointer to int type how about int &foo (void) what does it return? Thank a lot!

Nov 25, 2013 · It is a pointer to function that returns int* and accepts int* and pointer to function that returns int* (and accepts undefined number of parameters; see comments).

Jun 16, 2013 · -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 …

int* i, int * i, int*i, and int *i are all exactly equivalent. This stems from the C compiler (and it's compatible C like systems) ignoring white space in token stream generated during the process …

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 …

Dec 29, 2022 · For INT and other numeric types that attribute only specifies the display width. See Numeric Type Attributes in the MySQL documentation: MySQL supports an extension for …

Jan 28, 2012 · 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 …

Dec 17, 2013 · │int││int││int││int││int││int││ └───┘└───┘└───┘└───┘└───┘└───┘└┄ Of course, there's no reason they can't all …

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 …

Mar 29, 2012 · 3 Every expression in C or C++ has a type, a value, and possible side-effects. int i; ++i; The type of ++i is int. The side-effect is to increment i. The value of the expression is the …