Define and inline

1. We write an operation that asks for a number of squares,

   #define Macro definition is implemented as follows:

   #define SQUARE1 (X) x*x

   inline Inline functions are implemented as follows:

   inline int SQUARE2 (int X) {return x*x;}

   END

Step 2--define the required variables

1. 1

   First define the required variables:

   int a=5,b=5;

   int s1=0,s2=0;

   Define A and B values of the same, where a generation into #define test, B-generation into the inline test;

   S1 stores #define results, S2 stores inline results.

   END

Step 3--test # define and inline

2. 1

   Test 1: Write the following code:

   s1 = SQUARE1 (a);

   cout<< "s1=" <<S1<<ENDL;

   s2 = SQUARE2 (b);

   cout<< "s2=" <<S2<<ENDL;

    

   Result output:

   S1=

   S2=

   You can see the same output , analysis:

   S1 = SQUARE1 (a); Equivalent to

   s1 = a*a;

   So S1 is equal to the square of A;

    

   s2 = SQUARE2 (b); The equivalent of

   makes the SQYARE2 () function call First, B (5) as the argument pass, the function returns the value of 5*5, and then assigns the value to S2, so S2 is also equal to +

4. 2

   Test 2: Write the following code:

   s1 = SQUARE1 (a+b);

   cout<< "s1=" <<S1<<ENDL;

   s2 = SQUARE2 (a+b);

   cout<< "s2=" <<S2<<ENDL;

    

   Result output:

   S1=

   S2=

   You can see that the output is not the same , analysis:

   S1 = SQUARE1 (a+b); Equivalent to

   s1 = a+b*a+b;

   The

   can see that # define is simply replaced, and because the operator has precedence, it gets

   S1 = 5+25+5=35;

    

    

   s2 = SQUARE2 (a+b); Equivalent to

   First SQYARE2 () function call, a+b result 10 is passed as argument, the function returns the value of 10*10, and then assigns the value to S2, so S2 is also equal to 100.

    

   Here involves the program Order Point knowledge, corresponding to this is a # define only simple substitution, the entire SQUARE1 (a+b) is followed by sequential points, For the SQUARE2 (a+b) defined by the inline, there are two sequential points, the first after A+b, and the second after the entire SQUARE2 (a+b) , which means that the program will perform the a+b operation first (Order point 1). The resulting results are then involved in the SQUARE2 () internal code operation (Order point 2).

6. 3

   Some might say that the following improvements can be taken to address the problem:

   #define SQUARE1 (x) ((x) * (x )

   It is true that the above problem is solved, but will there be other problems?

   Look at the following Test 3:

8. 4

   Test 3: Write the following code:

    

   S1 = SQUARE1 (a++);

   cout<< "s1=" <<S1<<ENDL;

   cout<< "a =" <<A<<ENDL;

   s2 = SQUARE2 (b++);

   cout<< "s2=" <<S2<<ENDL;

   cout<< "b =" <<b<<endl;

    

   Result output:

   S1=

   A = 7

   S2=

   B = 6

   You can see the output of both is different , analysis:

   S1 = SQUARE1 (a++); is equivalent to

   s1 = a++*a++; because the suffix + + is incremented after use, so

   S1 = 5*5 = Two

   Then an increments of a=7

    

   s2 = (= = = = = + = =). SQUARE2 (b++); Equivalent to

   First SQUARE2 () function call, also because the suffix + + after the first use of the principle of increment, B, the initial value of 5 as an argument passed, the function returns the value of 5*5, and then assign the value to S2, so S2 equals;

   Then b++ get b=6

    

   Referring to the knowledge of the order Point described earlier, you will see that in this case #define increments the argument two times, and inline increments the argument one time , the problem persists even in the third step of the improvement, as described in the following test

10. 5

    The measured results are as follows:

    #define SQUARE1 (X) x*x

12. 6

    #define SQUARE1 (x) ((x) * (x))

    You can see that the problem with test 2 is resolved, but the test 3 problem persists

    END

Precautions

• If you are using C-language macros to perform similar functions, you should consider converting them to C + + inline functions

Define and inline