Would anyone please analyze the array problem to me please?

[kun niu]

Dear all,
I've got the following code:
int main()
{
int *a[6][3];
cout<<(a[5]-a[2])<<endl;
cout<<a[5]<<" "<<a[2]<<endl;
return 0;
}
But the output is like this:
9
0xbff44854 0xbff44830
I know that 9 comes from (5-2)*3.
But why is the output here not 36 but 9?

Thanks for any attention and appreciate your answer in advance.

 

[Sjouke Burry]

Dear all,
I've got the following code:
int main()
{
int *a[6][3];
cout<<(a[5]-a[2])<<endl;
cout<<a[5]<<" "<<a[2]<<endl;
return 0;
}
But the output is like this:
9
0xbff44854 0xbff44830
I know that 9 comes from (5-2)*3.
But why is the output here not 36 but 9?

Thanks for any attention and appreciate your answer in advance.

You have not initialized anything, so what do you expect??
Those variables you used, contain random trash.

 

[kun niu]

Dear all,
I've got the following code:
int main()
{
int *a[6][3];
cout<<(a[5]-a[2])<<endl;
cout<<a[5]<<" "<<a[2]<<endl;
return 0;
}
But the output is like this:
9
0xbff44854 0xbff44830
I know that 9 comes from (5-2)*3.
But why is the output here not 36 but 9?

Thanks for any attention and appreciate your answer in advance.

Given the following declaration:

int *a;

With the given declaration, ++a does not increment the address value of 'a'
by 1, but rather by sizeof(int), so that the memory address "++a" points to
the next, following, int in memory.

a+1 returns the same value: not the memory address of 'a' plus 1, but a
memory address of the next int, after 'a'.

Similarly: given two pointers:

int *a, *b;

Presuming that both a and b are pointing to elements of the same array,
the expression (b-a) does not result in the difference between the memory
addresses of b and a, but that difference divided by sizeof(int).

This is so that:

int c=b-a;

This gives the number of ints between a and b, not the number of bytes
between a and b, so that "a+c" returns the same pointer as 'b', or, in other
words:

(b-a) + a == b

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Thank you for your detailed explanation.
The answer is really what I need.