例如: Sizeof (char*)返回4。还有int* long long*,我试过的所有方法。有什么例外吗?
当前回答
The size of the pointer basically depends on the architecture of the system in which it is implemented. For example the size of a pointer in 32 bit is 4 bytes (32 bit ) and 8 bytes(64 bit ) in a 64 bit machines. The bit types in a machine are nothing but memory address, that it can have. 32 bit machines can have 2^32 address space and 64 bit machines can have upto 2^64 address spaces. So a pointer (variable which points to a memory location) should be able to point to any of the memory address (2^32 for 32 bit and 2^64 for 64 bit) that a machines holds.
由于这个原因,我们看到指针的大小在32位机器中是4字节,在64位机器中是8字节。
其他回答
从技术上讲,C标准只保证sizeof(char) == 1,其余的取决于实现。但在现代x86架构(例如Intel/AMD芯片)上,这是相当可预测的。
You've probably heard processors described as being 16-bit, 32-bit, 64-bit, etc. This usually means that the processor uses N-bits for integers. Since pointers store memory addresses, and memory addresses are integers, this effectively tells you how many bits are going to be used for pointers. sizeof is usually measured in bytes, so code compiled for 32-bit processors will report the size of pointers to be 4 (32 bits / 8 bits per byte), and code for 64-bit processors will report the size of pointers to be 8 (64 bits / 8 bits per byte). This is where the limitation of 4GB of RAM for 32-bit processors comes from -- if each memory address corresponds to a byte, to address more memory you need integers larger than 32-bits.
The size of the pointer basically depends on the architecture of the system in which it is implemented. For example the size of a pointer in 32 bit is 4 bytes (32 bit ) and 8 bytes(64 bit ) in a 64 bit machines. The bit types in a machine are nothing but memory address, that it can have. 32 bit machines can have 2^32 address space and 64 bit machines can have upto 2^64 address spaces. So a pointer (variable which points to a memory location) should be able to point to any of the memory address (2^32 for 32 bit and 2^64 for 64 bit) that a machines holds.
由于这个原因,我们看到指针的大小在32位机器中是4字节,在64位机器中是8字节。
只是已经发布的列表中的另一个例外。在32位平台上,指针可以占用6字节,而不是4字节:
#include <stdio.h>
#include <stdlib.h>
int main() {
char far* ptr; // note that this is a far pointer
printf( "%d\n", sizeof( ptr));
return EXIT_SUCCESS;
}
如果你用Open Watcom编译这个程序并运行它,你会得到6,因为它支持的远指针由32位偏移量和16位段值组成
如果您正在为64位机器编译,那么它可能是8。
即使是在普通的x86 32位平台上,你也可以得到不同大小的指针,试试这个例子:
struct A {};
struct B : virtual public A {};
struct C {};
struct D : public A, public C {};
int main()
{
cout << "A:" << sizeof(void (A::*)()) << endl;
cout << "B:" << sizeof(void (B::*)()) << endl;
cout << "D:" << sizeof(void (D::*)()) << endl;
}
在Visual c++ 2008中,指向成员函数的指针的大小分别为4、12和8。
Raymond Chen在这里讲过。