A discussion post on Array pointers in the csdn Forum

This is a good question. I sorted it out for details:

Http://topic.csdn.net/u/20091214/10/da5c0155-6055-4ef5-8cb6-cad8fc72136c.html? Seed = 1804958680 & R = 61913848 # r_61913848

 

Problem:

# Include "stdio. H"
Void main ()
{
Int y [4] [3] = {1, 1}, {1, 1}, {1, 1}, {1, 1 }};
Printf ("% d/N", * (Y + 2) + 2-* y, & Y [2] [2]-y [0], sizeof (* (Y + 2) + 2), sizeof (* y ));
Int Z [4] [3];
Printf ("% d", * (Z + 2) + 2-* z, & Z [2] [2]-Z [0]);
}

The VC compiler is used. An integer occupies 4 bytes and the output result is unknown.
8 press any key
Continue) the first two 8 values are equivalent. In this case, the integer value occupies 4 bytes, which should be 32 rather than 8. Why is it 8? The following results are 4 and 12, I don't know how to understand it,
They all indicate the length of the address. The address in VC is 4 bytes and should be equal. Why is the result different?

 

 

 

Answer:

The Y type is int [4] [3].
In the expression y involved, the value of Y is converted to the INT (*) [3] type.
Y [I] I =, the type of 2 and 3 is int [3],
In the expression y [I], the value of Y [I] is converted to the int * type.

The expression * (Y + 2) + 2-* Y is equivalent to Y [2] + 2-y [0].
The types of Y [2] and Y [0] are converted to int *, and there are 6 int values between them. Therefore, Y [2]-y [0] = 6, 2 is 8.

& Y [2] [2]-y [0]
The Y [2] [2] type is int, And the & Y [2] [2] type is int *, which is 8 different from Y [0], so it is 8

Sizeof (* (Y + 2) + 2) is equivalent to the type of sizeof (Y [2] + 2) y [2] in the formula instead of int *, so the type of Y [2] + 2 is int *, which is 4

In sizeof (* y) * y, convert it to int (*) [3], so * Y is int [3]. Therefore, sizeof (* Y) is 12.
In this case, * Y is Y [0] and the type is int [3]. Therefore, the result is 12.