C ++定义预处理器

Question

I am learning C++ and right we are covering preprocessors but I am trying to solve a question from a quiz which I has confused me a bit or a lot.. I tried to worked out by my own before running the program.. and my output was..

System started...
Data at 2 is: 27 28 29 30
Data at 1 is: 23 24 25 26
The data is: 19

I checked the program in Xcode to see if my output is right but the right output is the next one:

System started...
Data at 1 is: 0 0 0 19
Data at 0 is: 7 0 0 0
The data is: 19 0 0 0

This is the code...

#include <iostream>

namespace test{
#define COMPILE_FAST
#define PRINT_SPLIT(v) std::cout << (int)*((char*)(v)) << ' ' << \
(int)*((char*)(v) + 1) << ' ' << (int)*((char*)(v) +2) << ' ' << \
(int)*((char*)(v) + 3) << std::endl

    typedef unsigned long long uint;
    namespace er{
        typedef unsigned int uint;
    }
    void debug(void* data, int size = 0){
        if(size==0){
            std::cout << "The data is: ";
            PRINT_SPLIT(data);
        } else {
             while(size--){
                std::cout << "Data at " << size << " is: ";
                char* a = (char*)data;
                PRINT_SPLIT((a + (4+size)));
            }
        }
    }
}// End of Test namespace...

int main(){
    test::uint a = 19;
    test::er::uint b[] = {256,7};
    std::cout << "System started..." << std::endl;
    test::debug(b,2);
    test::debug(&a);
    std::cout << "Test complete";
    return 0;
}

My big doubt or what I actually don't understand is whats going on here in this preprocessor because clearly for what I did its totally wrong...

#define PRINT_SPLIT(v) std::cout << (int)*((char*)(v)) << ' ' << \
(int)*((char*)(v) + 1) << ' ' << (int)*((char*)(v) +2) << ' ' << \
(int)*((char*)(v) + 3) << std::endl

if someone can be so nice and give me a brief explanation I will extremely appreciate it.

Answer 1

The macro prints the values (as ints) of 4 consecutive bytes. It allows you to see how a 4 byte int is layed out in memory.

Memory contents, by byte, look like this (base10):

0x22abf0:       0       1       0       0       7       0       0       0
0x22abf8:       19      0       0       0       0       0       0       0

• 0 1 0 0 is 256, ie b[0]
• 7 0 0 0 is 7, ie b[1]
• 19 0 0 0 0 0 0 0 is 19, ie a

The sizeof(a) is different than the sizeof(b[0]) because there are 2 different typedefs for uint . Namely, test:uint and test::er::uint .

The address of a is greater than the address of b[] even though b is declared after a because the stack is growing downwards in memory.

Finally, I would say the output represents a defective program because the output would more reasonably be:

System started...
Data at 1 is: 7 0 0 0
Data at 0 is: 0 1 0 0
The data is: 19 0 0 0

To get that output the program needs to be changed as follows:

         while(size--){
            std::cout << "Data at " << size << " is: ";
            int* a = (int*)data;
            PRINT_SPLIT((a + (size)));