为什么4而不是1呢INT指针'++'增量？

问题描述：

的指针的值是可变的地址。为什么价值 INT指针整型指针增加1后，增加了4个字节。

Value of a pointer is address of a variable. Why value of an int pointer increased by 4-bytes after the int pointer increased by 1.

在我看来，我觉得指针（变量的地址）的价值指针递增后仅1个字节的增加。

In my opinion, I think value of pointer(address of variable) only increase by 1-byte after pointer increment.

测试code：

int a = 1, *ptr;
ptr = &a;
printf("0x%X\n", ptr);
ptr++;
printf("0x%X\n", ptr);

期望的输出：

0xBF8D63B8
0xBF8D63B9

实际输出的：

0xBF8D63B8
0xBF8D63BC

修改

另外一个问题 - 如何一个参观4个字节的 INT 占用一个

Another question - How to visit the 4 bytes an int occupies one by one?

答

当你增加一个 T * ，它移动的sizeof（T）字节^†这是因为它没有任何意义，以移动任何其他值：如果我指着一个 INT 这是4字节大小，例如，你会递增不到4离开我？的部分 INT 与其他一些数据混合。无厘头

When you increment a T*, it moves sizeof(T) bytes.^† This is because it doesn't make sense to move any other value: if I'm pointing at an int that's 4 bytes in size, for example, what would incrementing less than 4 leave me with? A partial int mixed with some other data: nonsensical.

在内存试想一下：

    [↓      ]
[...|0 1 2 3|0 1 2 3|...]
[...|int    |int    |...]

当我增加了指针，它更有意义？这样的：

Which makes more sense when I increment that pointer? This:

            [↓      ]
[...|0 1 2 3|0 1 2 3|...]
[...|int    |int    |...]

或者这样：

      [↓      ]
[...|0 1 2 3|0 1 2 3|...]
[...|int    |int    |...]

最后实际上并不指向一个任何种类的 INT 的。（从技术上讲，然后，使用该指针UB.)

The last doesn't actually point an any sort of int. (Technically, then, using that pointer is UB.)

如果您的真正的要移动一个字节，递增的char * ：的字符大小code>总是之一：

If you really want to move one byte, increment a char*: the size of of char is always one:

int i = 0;
int* p = &i;

char* c = (char*)p;
char x = c[1]; // one byte into an int

†这样做的必然结果是，你不能增加无效* ，因为无效是一个不完整的类型。

†A corollary of this is that you cannot increment void*, because void is an incomplete type.

为什么4而不是1呢INT指针'++'增量？

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