以下代码使用 arrays、指针等打印的内容

Question

I have problem solving this problem, so if anyne had a similar problem it would help me a lot.

short y[2][3]={{0123},{0x12345}},*p=y[1];
printf("01:%x\n", y);
printf("02:%x\n", p);
printf("03:%x\n", sizeof(y));
printf("04:%x\n", sizeof(y[0]));
printf("05:%x\n", sizeof(&y[0]));
printf("06:%x\n", sizeof(*p));
printf("07:%x\n", sizeof(p++));
printf("08:%x\n", *p++);
printf("09:%x\n", *p);
return 0;

Can anyone explain to me why the printout is like this?

01:61ff10
02:61ff16
03:c
04:6
05:4
06:2
07:4
08:2345
09:0

My opinion:

01:Prints the address where the array y begins.
02:Prints the address of the pointer, which points to the second element of the array. Since we have 2 * 3 elements that are of type short, each subsequent element of the zero element will increase by 6.
03:Since we have 2 * 3 elements, which is equal to 6, but the elements of the type are short, so it will print hexadecimal c
04:the number of elements in the zero position is 3, but they are of the short type, so it prints 6
05:prints the sizeof addresses of the first element of the array which is 4
06:I don't know why it prints 2 here
07:Prints the sizeof of the pointer address which is 4, it will increase after printing
08:I do not understand
09:I do not understand

Can anyone explain why it prints like this?

Answer 1

OK, let's see:

• #01: The address of y .
• #02: The value of p , which holds the address of y[1] , which is the second element of type short[3] . The size of a short is apparently 2 on your system, so the offset to #01 is 6.
• #03: The size of the array y , 2 * 3 * sizeof (short) give 12, in hex c .
• #04: The size of the element y[0] , which is of type short[3] . 6, as you found.
• #05: The size of the address of y[0] , and apparently the size of an address is 4 on your system.
• #06: The size of the object that p points to. This is a short , so 2.
• #07: The size of the expression p++ , which is an address, so 4. And no, p is not incremented, since the expression is not evaluated.
• #08: The value of the object that p points to, which is y[1][0] . Since the initializing value of 0x12345 is an int too big to be stored in a short , it is truncated to 0x2345 . After reading the value, p is incremented.
• #09: The element p points to, which is y[1][1] . It was initialized to 0.

Notes:

You should have got warnings from your compiler:

• The mentioned initializer is truncated.
• The format for pointers/addresses is %p .
• The type of the result of sizeof might not match the format %x .

You should take warnings seriously, they are always a hint that you most probably made an error.

Answer 2

N6) Sizeof(*p) is size of datatype pointed by p. p is pointer to short: so 2 bytes.

N8) p is pointer to short, it`s pointing to array element y[1][0]. y[1] and y[1][0] have the same memory address.

All array elements are short, 0x12345 truncates to 0x2345 upon array initialisation. So output is 2345. Also, p++ increases pointer value to point to next short y[1][1].

N9) Because of p++ in step N8, pointer now points to y[1][1], which was initialised to 0 (by default, because init value not provided) - output is 0.