I got a new title, just I didn't want to stick it again, so I copied it.
Give everyone a reference, don't cause everyone to pay attention to. Thank you!
Sender: Reaking (Reaking), Letter Area: Office
Title: RE: several interview questions
Sending station: Gulang listening to Tao (Fri Nov 16 17:19:03 2001), transfer
Faint ... still gives a reason.
[Mentioned in the masterpiece of Reicking (Release):]
: There is a little difficulty. I made a reference answer, the reason is not busy.
: [In the masterpiece of Tsky (Sky):]
:: Send a letter: Drinking water (11:53:18, November 16, 2001), transfer
: : 1. INT i = 5, J;
:: J = ( i) * (i -);
:: After running i = ______, j = _________
: I == 5; j == 36;
We know i first calculate i in participating in expression, and I - just the opposite.
Original equivalent (j = ( i) * i) && (i -)
(j == 6 * 6) && (i == 6-1)
: : 2. INT A [] = {0, 1, 2, 3, 4};
:: Int * p [] = {A, A 1, A 2, A 3};
::: Int ** pp = P;
:: Assumption a = 8239150 (can't remember, probably), P = 8239168
:: * (PP ) - a =?
: 2
The third line tells us PP = P;
So PP is equivalent to P ;
And P is equivalent to P 1; equivalent to & p [1];
So * (PP ) is equivalent to * (& P [1]); equivalent to P [1];
P [1]? Equal to A 1;
So the original equivalent price (A 1) -a;
Considering that A and A 1 are plastic addresses, accounting for 2 bytes;
So the answer is 2.
:: 3. INT A [100], I, * P = a;
::: For (i = 0; i <100; i ) a [i] = i;
::: For (i = 0; i <100; i )
:: {
:: * P = a [i];
:: G (& p);
::};
::: For (i = 0; i <100; i )
:: {
:: Printf ("% D", A [i]);
::};
:: Void g (int ** p);
:: {
:: ** p ;
:: * P ;
::}
: 991234567 ... 99
: Ie a [0] == 99; a [i] = i; where 1 <= i <= 99;
priority is higher than *, so * p is equivalent to * (p );
Therefore, the only function of the function VOID G (int ** P) is P from the self-increment (P );
There is no impact on the content points to the argument.
In the cycle of the main () function, when i == 99, * p == a [0] == a [99] == 99; other A [i] has no change.
So the answer is what I said above.
:: 4. The role of the following functions:
::: Fun (char * a, char * b)
:: {
::: While (* a = * b )
:: {};
:: Return;
::}
: String copy function.
This is a simple destiny, you know what is going on less than 2 seconds.
* A is equivalent to * (A ), so the original is equivalent to (* (A )) = (* (b ));
A character copy.
The condition of the cycle is * a == '/ 0', ie * b == '/ 0';
Is the character string copy function is still not obvious?
