错误：从“void *”转换为“int”失去精度 [英] error: cast from 'void*' to 'int' loses precision

问题描述

我有一个原型 void * myFcn（void * arg）的函数，它用作pthread的起始点。我需要将参数转换为一个int供以后使用：

  int x =（int）arg; 
  

编译器（GCC版本4.2.4）返回错误：

  file.cpp：233：error：从'void *'转换为'int'失去精度
   
解决方案
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< cstdint> 来定义它。
 
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;
The compiler (GCC version 4.2.4) returns the error:
file.cpp:233: error: cast from 'void*' to 'int' loses precision
What is the proper way to cast this?
解决方案
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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