我应该如何处理“从”void *“转换为”int“失去精度”当在64位机器上编译32位代码？ [英] How should I handle "cast from ‘void*’ to ‘int’ loses precision" when compiling 32-bit code on 64-bit machine?

问题描述

我有一个包在32位机器上编译和工作正常。我现在试图让它在64位机器上编译，并发现以下错误 -

I have a package that compiles and works fine on a 32-bit machine. I am now trying to get it to compile on a 64-bit machine and find the following error-

 error: cast from ‘void*’ to ‘int’ loses precision

有没有编译器标志来抑制这些错误？

Is there a compiler flag to suppress these errors? or do I have to manually edit these files to avoid these casts?

推荐答案

问题是，在32bits中， int （一个32位整数）将保存一个指针值。

The issue is that, in 32bits, an int (which is a 32bit integer) will hold a pointer value.

当移动到64位时， - 它不够大，不能容纳64位指针。 intptr_t 类型是为此而设计的。

When you move to 64bit, you can no longer store a pointer in an int - it isn't large enough to hold a 64bit pointer. The intptr_t type is designed for this.

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