从无符号字符*转换为无符号int

When I do the following, I get this error:

当我做以下操作时，我得到这个错误:

../src/Sample.cpp:19: error: cast from \u2018UINT8*\u2019 to \u2018UINT8\u2019 loses precision

. . / src /样品。错误:从\u2018UINT8*\u2019到\u2018UINT8\u2019失去精度

#include <iostream>
using namespace std;

typedef unsigned char UINT8;
typedef unsigned int UINT32;

#define UNUSED(X) X=X

int main() {
    UINT8 * a = new UINT8[34];
    UINT32 b = reinterpret_cast<UINT8>(a);

    UNUSED(b);

    return 0;
}

How would I go about solving this. Keep in mind I am not trying to convert string to unsigned long, rather converting char* (the ADDRESS value ) to int.

我该怎么解决这个问题呢。请记住，我并不是要将字符串转换为无符号长，而是要将char*(地址值)转换为int。

Thanks

谢谢

Solution:

解决方案:

Turns out that this problem has to do with the pointer size. On a 32 bit machine, the pointer size is 32 bit, and for 64 bit machine is of course 64. The above wont work on a 64 bit machine, but will on a 32 bit machine. This will work on a 64 bit machine.

事实证明这个问题与指针大小有关。在32位机器上，指针大小是32位，64位机器当然是64位。上述方法不适用于64位机，但适用于32位机。这将在64位机器上工作。

#include <iostream>
#include <stdint.h>

using namespace std;

typedef  uint8_t UINT8;
typedef int64_t UINT32;

#define UNUSED(X) X=X

int main() {
    UINT8 * a = new UINT8[34];
    UINT32 b = reinterpret_cast<UINT32>(a);
    UNUSED(b);

    return 0;
}

3 个解决方案

#1

Assuming that sizeof(int) == sizeof(void*) you can use this code to convert:

假设sizeof(int) = sizeof(void*)，您可以使用此代码进行转换:

int b = *reinterpret_cast<int*>(&a);

or variations thereof. I think static_cast will work too.

或变化。我认为static_cast也可以。

Of course a has to be l-value (able to be assigned to) to get the address of it. For non-l-values, you need to use a function and the good old union trick will work:

当然a必须是l-value(能够被分配到)才能得到它的地址。对于非l值，您需要使用一个函数，好的老联合技巧将会奏效:

int Pointer2Int (void* p)
{
    union { void* p; int i; } converter;
    converter.p = p;
    return converter.i;
}

#2

This isn't a good idea, but the line should read:

这不是一个好主意，但这句话应该是:

UINT32 b = reinterpret_cast<UINT32>(a);

reinterpret_cast takes the type of the destination type as the template parameter.

reinterpretation t_cast将目标类型的类型作为模板参数。

With the correct type g++ will tell you that it isn't a good idea:

正确的类型g++会告诉你这不是个好主意:

error: invalid cast from type ‘char’ to type ‘int’

see Cthutu's answer for a more correct approach.

请参阅Cthutu的答案，以获得更正确的方法。

#3

An option would be to :

一种选择是:

int *a; char b[64]; int c;

int *;字符b[64];int c;

a = (int*) malloc (sizeof(int));

sprintf(b, "%d", a);

sprintf(b,“% d”);

c = atoi(b);

#1