getting the address value without word alignment?

I have this question:

suppose that the size of "char" is one byte and the size of "int" and "float" is 4 bytes.

struct Employee
{
int age;
char sex;
char name[25];
    float info[3][10];
};
Employee x;

what is the address of "x.info[2][5]" without word alignment, supposing that t开发者_如何学JAVAhe adress of "x" is 1000 ?

note: this is not a homework but i had an exam and i answered this wronge. I really wold like to know the right answer.

thanx

I am not sure about other languages, but in C and C++ this is how a 2-D array will be stored in the memory:

Your float info[3][10] has 3*10 = 30 floats in the memory.

If the first memory location of your 2-D array is 1000 i.e &info[0][0] then your &info[2][5] will be 1000 + sizeof(float) * (2*10 + 5) = 1100 i.e first_memory_location + sizeof(data_type) * (row_no * total_no_columns + column_no).

Have a look at this link to understand the concept better

EDIT: The same calculation holds good for 2-D arrays inside a struct.

In your case, there is an int, a char and a char array of size 25. So your &info[0][0] = 1029

First_memory_location + sizeof(float) * ( row_no * total_no_columns + column_no)

1029 + 4 *(2*10 + 5) = 1129

[I will suppose that X in your question is noise from an alien spacecraft]

Address &info[2][5] is then info[2]+5 = (char*)info+ sizeof(float)*2*10+5*sizeof(float) == (char*) info + 100

sizeof(float)==4

I think in this case you need to add the memory space of all variables together in order, not just the address space of x.

So if the address of x is 1000, the address you want would be at

1000
+sizeof(int)         // age
+sizeof(char)        // sex
+(sizeof(char)*25)   // name
+(sizeof(float)*2*10) + sizeof(float)*5   // info[2][5]

Assuming usual sizes, that would be 1000 + 8 + 1 + 25 + 120 = 1154