Upgrade from C to C ++ to c ++

Upgrade from C to C ++ to c ++

 

• Const in CIt's just a "Read-Only variable"Is not a constant in the true sense,The pointer can be used to change it, as shown below:

1. 1 # include <stdio. h> 2 int main () 3 {4 const int a = 0; // declare that the read-only variable a is 05 int * p = (int *) &; // point a pointer to the space of a 6 * p = 100; // change the content of a's bucket 7 printf ("a = % d", ); // a is changed to 1008 return0; 9}

    

   


Const is a constant in the true sense in C ++, Cannot be changed by pointer

1. 1 # include <iostream> 2 usingnamespace std; 3 int main () 4 {5 const int a = 0; // declare that the read-only variable a is 0 6 int * p = (int *) & a; // give a pointer to the space of a 7 * p = 100; 8 cout <a; 9 return0; 10}

   1 constint a = 10; 2 constint B = 20; 3 int c [a + B];

    

   In C, the above Code reports the error "cannot determine the length of the array", because in C, const is only a read-only variable, essentially a variable, with storage space, of course, an error will occur when defining the length of an array using variables. However, it can be compiled in C ++, because C ++ strengthens const, making const a constant in the true sense.
   • Difference between const and # define
   # Define only replaces common text before compilation (preprocessing) without scope check!

   1. 1 # include <iostream> 2 usingnamespace std; 3 int f () {4 cout <B <endl; 5} 6 int g () {7 # define N 3 // only use N = 3 8 constint a = 100 in this function; // you want to use a = 100 9} 10 # define B 111 int main () 12 {13 cout <N <endl; 14 f (); 15 return0; 16}

       

      

   In code 4th, the Error "[Error] 'B' was not declared in this scope" occurs. B is not defined. Note # When a simple text is replaced before define compilation, only replace all the items under # define, and those located on # define are not replaced. Const is handled by the compiler and has the scope check and type check.

   1. 1 # include <iostream> 2 usingnamespace std; 3 int f () {4 constint a = 100; // only the scope of a = 100 is expected to be f () in 5} 6} 7 int main () 8 {9 cout <a <endl; // error, no definition of a, because a scope is only in f () 10 return0; 11}

       

      

   # Define can be used # undef to limit the scope, but it is a little troublesome:

   1. 1 int g(){2 #define3 int a=10;4 #undef5 }

       

    
   • C ++ enhances function types and Parameters
   The three-object operation in C ++ can return variables (provided that a and B cannot be constants! (A <B? ) = 50; this is wrong at first glance), but not in C (three orders operator (a <B? A: B) First, judge whether the left side of the question mark is true. If it is true, execute a; otherwise, execute B)
For example:

1. 1 int main()2 {3 int a=1;4 int b=100;5 (a<b?a:b)=50;6 printf("%d %d",a,b);7 }

    

However, the above Code in C may fail, and the three-object operator cannot return variables.