[請益] 一個小問題

[alleternal (鬆餅)]

int main()
{
   double a[10];
   double* b = new double;
   double* c = new double[10];
   double** d = new double*[10];

   cout << "a   = " << a << endl;
   cout << "a+1 = " << a+1 << endl;
   cout << "b   = " << b << endl;
   cout << "b+1 = " << b+1 << endl;
   cout << "c   = " << c << endl;
   cout << "c+1 = " << c+1 << endl;
   cout << "d   = " << d << endl;
   cout << "d+1 = " << d+1 << endl;
}

Some of what I said in the class were wrong --- the value of “c+1” is the
address of “c” plus the size of the object (i.e. 8 Bytes), NOT the size of
a pointer (i.e. 4 Bytes).

So for variables “a”, “b”, and “c”, their “+1” addresses will be
8 Bytes more. For “d”, it’s 4 Byte.

A sample output for the above code is:

a   = 0xfee6ef20
a+1 = 0xfee6ef28
b   = 0x97a7008
b+1 = 0x97a7010
c   = 0x97a7018
c+1 = 0x97a7020
d   = 0x97a7070
d+1 = 0x97a7074

關於上面這個:
1. b和c不都是double pointer, 按照老師說的, b+1 c+1的效果不是應該是
   b+1=0x97a7016(產生的object是double array? double的object是8 bytes)
   c+1=0x97a7026  (理由也是同上)

2. 0xfee6ef20 這是位置我知道, 可是他有代表什麼意義?例如:編碼...

--
※ 發信站: 批踢踢實業坊(ptt.cc) 
◆ From: 59.117.208.42
→ alleternal:double** d = new double*[10]; 這個d是什麼樣的data
→ alleternal:type? 後面new出來的又是什麼東西?產生在哪裡?
推 admitted:ㄟ這個是16進位的表示法，不是十進位
推 ric2k1:  推~~~ 是16進位
→ ric2k1:0xfee6ef20 是 stack memory 中 local object 的 address