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First , const is the same as #define
const constants in C ++ are similar to macro definitions
const int c = 5 ≈ #define c 5
const is a means to replace #define.
Procedure one:
intmain()
{
const int a=10;
printf("a=%d\n",a);
return 0;
}
Procedure two:
#define a 10
intmain()
{
//const int a=10;
printf("a=%d\n",a);
return 0;
}
The results of these two programs are the same, both define a constant a, note that there is no semicolon at the end of #define. The following two programs also illustrate this:
#define a 10
#define b 10
intmain()
{
int arr[a+b];
system("pause");
return 0;
}
Compilation succeeded!
#define a 10
#define b 10
intmain()
{
int arr[a+b];
return 0;
}
Compilation succeeded!
Second , the difference
Look at the program:
Now a is a macro definition, we know the macro definition, the preprocessing compiler will replace wherever the variable a is, that is, replace a with 10. Therefore, the a defined in the function fun1 can be used in the function fun2, which means that the macro definition has no scope check . Run can pass.
Then if you want to limit the scope of a to the function fun1, you can use the "unload macro" or "cancel macro" #undef to achieve the purpose.
#undef a - here to cancel the macro definition of a;
#undef - cancel all macro definitions here.
Looking at the const scope check, we define the variable b in fun1, and its scope is limited to the fun1 function, which is not available in the fun2 function, you can cancel //printf("b=%d\n" ,b); comment, found a compile-time error!
Conclusion :
const constants in C ++ are not the same as macro definitions
const constants are handled by the compiler, providing type checking and scope checking;
Macro definitions are processed by the preprocessor, with pure text replacement.
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