[英]How do I cast a pointer to an int
I'm trying to store the value of an address in a non pointer int variable, when I try to convert it I get the compile error "invalid conversion from 'int*' to 'int'" this is the code I'm using:我试图将地址的值存储在非指针 int 变量中,当我尝试转换它时,我收到编译错误“从‘int*’到‘int’的无效转换”这是我正在使用的代码:
#include <cstdlib>
#include <iostream>
#include <vector>
using namespace std;
vector<int> test;
int main() {
int *ip;
int pointervalue = 50;
int thatvalue = 1;
ip = &pointervalue;
thatvalue = ip;
cout << ip << endl;
test.push_back(thatvalue);
cout << test[0] << endl;
return 0;
}
int
may not be large enough to store a pointer. int
可能不足以存储指针。
You should be using intptr_t
.您应该使用
intptr_t
。 This is an integer type that is explicitly large enough to hold any pointer.这是一个显式大到足以容纳任何指针的整数类型。
intptr_t thatvalue = 1;
// stuff
thatvalue = reinterpret_cast<intptr_t>(ip);
// Convert it as a bit pattern.
// It is valid and converting it back to a pointer is also OK
// But if you modify it all bets are off (you need to be very careful).
You can do this:你可以这样做:
int a_variable = 0;
int* ptr = &a_variable;
size_t ptrValue = reinterpret_cast<size_t>(ptr);
Why are you trying to do that, anyway you just need to cast, for C code :你为什么要这样做,无论如何你只需要为 C 代码进行转换:
thatvalue = (int)ip;
If your writing C++ code, it is better to use reinterpret_cast
如果你写 C++ 代码,最好使用
reinterpret_cast
我建议使用reinterpret_cast
:
thatvalue = reinterpret_cast<intptr_t>(ip);
I was able to use the C union statement to achieve what you were looking for.我能够使用 C union 语句来实现您想要的。 It will of course be compiler dependent, but it worked for me like you would presume it should (Linux, g++).
它当然依赖于编译器,但它对我有用,就像你认为它应该(Linux,g++)。
union {
int i;
void *p;
} mix;
mix.p = ip;
cout << mix.i << endl;
On my particular instance my int is 32 bit and the pointer is 48 bit.在我的特定实例上,我的 int 是 32 位,指针是 48 位。 When assigning the pointer, the integer value i will represent the lowest 32 bit of the pointer.
分配指针时,整数值 i 将代表指针的最低 32 位。
Since int *ip;
由于
int *ip;
is a pointer to an integer and int thatvalue = 1;
是一个指向整数和
int thatvalue = 1;
的指针int thatvalue = 1;
is an integer, assuming you want the value stored at the address pointed to by ip
assigned to thatvalue
, change thatvalue = ip;
是一个整数,假设要存储在由指向的地址的值
ip
分配给thatvalue
,变化thatvalue = ip;
to thatvalue = *ip;
到
thatvalue = *ip;
(note the addition of the dereference operator *
to access a value equivalent to the value at the pointer address ). (注意添加了取消引用运算符
*
以访问与指针地址处的值等效的值)。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.