The const analysis of the 3rd course after evolution

1. Const in the C language

(1) The const-modified variable is read-only , making the variable read-only, but essentially a variable . So not a true constant, it just tells the compiler that the variable cannot appear to the left of the assignment symbol.

(2) const-Modified local variable allocates space on stack , global variable allocates space in read- only storage

(3) const only useful at compile time , useless at run time

Const in the "programming experiment" C + +

#include <stdio.h>intMain () {Const intc =0;//The C language allocates memory for variable C    int* p = (int*) &c;//& only allocates memory for C C + +printf ("begin...\n"); *p =5;//the value in memory has already been changed to 5.printf ("C =%d\n", c);//the C language outputs 5 of the memory. //In C + +, values are taken from the symbol table (not memory)//so for 0.printf ("end...\n"); return 0;}

2. Const in C + +

(1) C + + gives precedence to const on the basis of C, and puts constants in the symbol table when a const declaration is encountered.

(2) If a constant is found during compilation, it is replaced directly by the value in the symbol table .

(3) If you find that the extern or & operator is used for const constants during compilation, the corresponding constants are allocated storage space . Note that although the C + + compiler may allocate space for const constants, it does not use the values in its storage space.

3. Comparing the const in C/

	C language	C++
Essence	Read-only variables	Constant
Allocating memory	Will be assigned	When you use the & operator to assign addresses to const constants Memory is allocated when Const constants are global and need to be used in other files

4. The difference between a const and a macro in C + +

	Const in C + +	Macro
Defined	const int c = 5;	#define C 5
Processing mode	Handled by the compiler , the compiler does type checking and scope checking	Processed by the preprocessor , just simple text substitution

"Programming Experiment" Const and macro

#include <stdio.h>voidf () {//macros are processed by precompilation, and the macros behind them work    #defineA 3Const intb =4;//scope is limited to the F function}voidg () {printf ("A =%d\n", a);//valid, as long as the macro definition can be used//printf ("b =%d\n", b);//the scope of illegal b is limited to the F function}intMain () {Const intA =1; Const intB =2; intArray[a + B] = {0};//C + + is legal because it thinks A and B are constants. //the const nature of C is still a variable, and the size of the array can only be constant .    inti =0;  for(i=0;i< (A + B); i++) {printf ("array[%d] =%d\n", I, array[i]);    } f ();    g (); return 0;}

5. Summary

(1) Unlike the C language, const in C + + is not a read-only variable

(2) const in C + + is a true constant

(3) C + + compiler may allocate space for const constants

(4) C + + is fully compatible with the syntax characteristics of const constants in C language

The const analysis of the 3rd course after evolution