[英]How much memory is required to represent an address?
Consider the following piece of code. 考虑以下代码。
int var;
cout << (long)&var;
My doubt is how do we know that long int has sufficient width to hold the memory location indicated by &var
. 我怀疑的是我们怎么知道long int有足够的宽度来保存
&var
指示的内存位置。 What if it is not sufficient? 如果还不够怎么办?
The full code which I am executing... 我正在执行的完整代码......
//: C03:YourPets2.cpp
// From Thinking in C++, 2nd Edition
// Available at http://www.BruceEckel.com
// (c) Bruce Eckel 2000
// Copyright notice in Copyright.txt
#include <iostream>
using namespace std;
int dog, cat, bird, fish;
void f(int pet) {
cout << "pet id number: " << pet << endl;
}
int main() {
int i, j, k;
cout << "Address size " << sizeof(&f) << endl;
cout << "Long size " << sizeof(long) << endl;
cout << "Intptr size " << sizeof(intptr_t) << endl;
cout << "f(): " << &f << endl;
cout << "f(): " << (long)&f << endl;
cout << "f(): " << (long long)&f << endl;
cout << "dog: " << (long)&dog << endl;
cout << "cat: " << &cat << endl;
cout << "bird: " << &bird << endl;
cout << "fish: " << (long)&fish << endl;
cout << "i: " << (long)&i << endl;
cout << "i: " << (long long)&i << endl;
cout << "j: " << (long)&j << endl;
cout << "k: " << (long)&k << endl;
} ///:~
The result which I am getting: 我得到的结果:
Address size 4
Long size 4
Intptr size 4
f(): 1
f(): 134514548
f(): 134514548
dog: 134521044
cat: 0x804a0d8
bird: 0x804a0dc
fish: 134521056
i: -1074729380
i: -1074729380
j: -1074729384
k: -1074729388
You don't. 你没有。 It's possible - if unlikely - for pointers to have larger storage requirements than any integer.
对于具有比任何整数更大的存储要求的指针,它是可能的 - 如果不可能的话。 If there is an integer type that is suitable then there will be a typedef for it
std::intptr_t
(and possible also std::uintptr_t
) defined in <cstdint>
(C++11 only). 如果有一个合适的整数类型,那么在
<cstdint>
定义的std::intptr_t
(以及可能的std::uintptr_t
)将有一个typedef(仅限C ++ 11)。
You can test for the presence of intptr_t
at the preprocessor stage by testing for the definedness of the macro INTPTR_MAX
(or INTPTR_MIN
) after #include <cstdint>
. 您可以通过在
#include <cstdint>
之后测试宏INTPTR_MAX
(或INTPTR_MIN
)的定义性来测试预处理器阶段是否存在intptr_t
。
If you just want to print a pointer value using std::cout
then you can cast to void*
(unnecessary for int*
but necessary for char*
) and use <<
directly without a cast to an integer type. 如果你只想使用
std::cout
打印一个指针值,那么你可以转换为void*
( int*
必需的,但char*
必需的)并且直接使用<<
而不使用强制转换为整数类型。
You can use an assert (or a different kind of check). 您可以使用断言(或不同类型的检查)。 The check should be in the form
检查应该在表格中
assert( sizeof( &var ) <= sizeof(long) );
assert(sizeof(&var)<= sizeof(long));
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