gcc将int *转换为枚举*错误 [英] gcc convert int * to enum * error

问题描述

#include <stdio.h>
typedef enum{
red=1,blue,green
}color;

void set_color(color *c)
{
  *c = blue;
}

int main()
{
  //int a = 3;
  //set_color(&a);//why this error?
  //set_color((color *)(&a));//It is right
  color c = red;
  set_color(&c); // this is right;
  return 0;
}

i使用gcc -S用set_color（（color *）（& a）查看汇编语言）和set_color相同（ & c），但为什么gcc不自动执行转换？

i use gcc -S to view the assemble language with set_color((color *)(&a)) and it is the same with set_color(&c),but why gcc does not do the convertion automatically?

推荐答案

因为你定义了一个类型。因此隐式铸造不起作用。请参阅： http://en.cppreference.com/w/cpp/language/implicit_cast [ ^ ]

Because you defined a type. Thus implicit type casting won''t work. See: http://en.cppreference.com/w/cpp/language/implicit_cast[^]

因为它试图阻止你犯错误，而你却没有意识到这一点。

当你写作

Because it is trying to prevent you making a mistake, and you not realizing it.
When you write

set_color((color *)(&a));

你明确地转换它 - 告诉编译器这是我想要发送的值，我知道我在做什么所以它让它通过。



写的时候

You are explicitly converting it - telling the compiler "this is the value I wanted to send, I know what I am doing" so it lets it through.

When you write

set_color(&a);

这可能是一个错误 - 你可能没有意识到a根本就不是一种颜色。因此它会引发错误，因此您可以在问题成为运行时问题之前解决问题并导致奇怪的错误。

It could be a mistake - you might not have realised that "a" was not a color at all. So it raises an error so you can fix the problem before it becomes a run time problem and causes an odd bug.

在C（和C ++）中，数据类型为''int''和''enum''部分是类型安全的。例如：

In C (and C++), the datatypes ''int'' and ''enum'' are partly typesafe. For example:

typedef enum {E0 = 0, E1 = 1} ENUMtype_t;

ENUMtype_t e = E0;  // legal
int i = 0;          // legal
i = E0;             // legal
e = 0;              // legal(!)
e = 1;              // error(!)
e = i;              // error
e = (ENUMtype_t) i; // legal

这篇关于gcc将int *转换为枚举*错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！