类型转换变量与其他的typedef

问题描述

  typedef结构{
  unsigned char型一，
  unsigned char型B，
  unsigned char型Ç
} type_a;typedef结构{
 unsigned char型E，
 unsigned char型F [2]
} type_b;type_a样本;
sample.a = 1;
sample.b = 2;
sample.c文件= 3;
 

我如何强制转换它，并赋予新的价值，如：

 样品=（type_b）样品; //语法错误
sample.f [1] ='A';
 

解决方案

您不能施放一种数据类型到另一种不兼容的数据类型。然而，存储器是为你打开。您可以通过以下方式访问它：

  typedef结构
{
  unsigned char型一;
  unsigned char型B：
  unsigned char型℃;
} type_a;typedef结构
{
 unsigned char型ê;
 unsigned char型F [2];
} type_b;type_a样本;
sample.a = 1;
sample.b = 2;
sample.c文件= 3;type_b * sample_b =（type_b *）（（无效*）及样品）;
 

试用自己访问 sample_b-＆GT; e 和 sample_b-方式＆gt;˚F键，看看会发生什么

typedef struct {
  unsigned char a,
  unsigned char b, 
  unsigned char c
}type_a;

typedef struct {
 unsigned char e,
 unsigned char f[2]
}type_b;

type_a sample;
sample.a = 1;
sample.b = 2;
sample.c = 3;

How do I typecast it and assign new value like:

sample = (type_b)sample; // syntax error
sample.f[1] = 'a';

解决方案

You can't cast one data type to another incompatible data type. However, the memory is open for you. You can access it as follows:

typedef struct
{
  unsigned char a;
  unsigned char b; 
  unsigned char c;
}type_a;

typedef struct
{
 unsigned char e;
 unsigned char f[2];
}type_b;

type_a sample;
sample.a = 1;
sample.b = 2;
sample.c = 3;

type_b *sample_b = (type_b *) ((void*) &sample);

Try out yourself accessing sample_b->e and sample_b->f and see what happens.

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