错误:从“void*"转换为“int"失去精度

error: cast from #39;void*#39; to #39;int#39; loses precision(错误:从“void*转换为“int失去精度)

问题描述

我有一个原型为 void* myFcn(void* arg) 的函数，它用作 pthread 的起点.我需要将参数转换为 int 以备后用:

I have a function with prototype void* myFcn(void* arg) which is used as the starting point for a pthread. I need to convert the argument to an int for later use:

int x = (int)arg;

编译器(GCC 4.2.4 版)返回错误:

The compiler (GCC version 4.2.4) returns the error:

file.cpp:233: error: cast from 'void*' to 'int' loses precision

投射这个的正确方法是什么?

What is the proper way to cast this?

推荐答案

您可以将其转换为 intptr_t 类型.它是一个 int 类型，保证足够大以包含一个指针.使用 #include 来定义它.

You can cast it to an intptr_t type. It's an int type guaranteed to be big enough to contain a pointer. Use #include <cstdint> to define it.

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