宏打印不正确的输出

Question

I'm learning C programming and this is my problem. I feel like I've learned the macro topic in C but I guess I'm not quite ready yet.

#define PGSIZE    4096 
#define CONVERT(sz)    (((sz)+PGSIZE-1) & ~(PGSIZE-1))

printf("0x%x", CONVERT(0x123456));

Here is the problem. My expected output is 0x100000000000 but it prints 0x124000 .

((sz)+PGSIZE-1) = (0x123456)+4096-1 
                = (0x123456)+(0x1000 0000 0000) - 1 //4096 is 2^12
                =  0x1000 0012 3456 - 1
                =  0x1000 0012 3455

~(PGSIZE-1) => ~(0x0111 1111 1111) = 0x1000 0000 0000

((sz)+PGSIZE-1) & ~(PGSIZE-1) = (0x1000 0012 3455) & (0x1000 0000 0000)
                              = 0x100000000000

But when I ran the program, it prints 0x124000.

What am I doing wrong?

Answer 1

You showed in the question:

((sz)+PGSIZE-1) => (0x123456)+4096-1 
                =(0x123456)+(0x1000 0000 0000) - 1 //4096 is 2^12
                =0x1000 0012 3456 - 1

You converted 4096 to a binary notation, but then treat it as a hexadecimal number. That won't work. If you want to keep the hexadecimal notation, that's:

((sz)+PGSIZE-1) => (0x123456)+4096-1 
                =(0x123456)+(0x1000) - 1
                =0x124456 - 1

Or converting both to binary, that's:

((sz)+PGSIZE-1) => (0x123456)+4096-1 
                =(0b1_0010_0011_0100_0101_0110)+(0b1_0000_0000_0000) - 1
                = 0b1_0010_0100_0100_0101_0110 - 1

Answer 2

The error is in your calculation. 2^12 is not 1000 0000 0000, but 0001 0000 0000 0000.

The weights of binary begin as 2^0 which is one so 2^12 comes at 13th position so 4096 is 0x1000 If you use this for your manual calculation you will get 0x124000 as your answer.

The below calculation also answers your doubt "how 0x124455 & 1000 becomes 0x124000? Does if automatically fill 1s to the front? Could you explain little more about it on the question?" in the comment of the previous answer.

4096 = 0x1000 
4096-1 => 0xfff => 0x0000 0fff
    ~(4096-1) is thus 0xfffff000

    Coming to the addition part in macro 
     (0x123456)+4096-1 
         =>0x123456+0x1000-1
         =>0x124456-1
         =>0x124455

    You result will be 0x124455 & 0xfffff000 which is 0x124000 which is the correct output