如何从一种类型转换到另一个在C

问题描述

我有以下code

#include <stdio.h>
#include<ctype.h>

typedef struct {
  int Type;
  int Type2;
}foo;

typedef struct {
  char cData[40];
}bar;

int main()
{
  bar b1;
  strcpy(b1.cData,"11");
  foo *f=(struct foo *)&b1;
  printf("Type is  %d \n",f->Type);
  return 0;
}

但我没有得到f中的指针类型1的值，而不是我得到了particuler结构的大小。

But i am not getting the value of type 1 in f's pointer , instead i am getting size of that particuler struct.

推荐答案

当我运行code（修正错误后），它打印12593.这是49 * 256 + 49 - 换句话说，11 作为一个整数（即49 ASCII 1）。所以没有错code，据我所看到的（除了内存布局的假设所指出的伯努瓦），所以我们需要知道你期望发生

When I run the code (after correcting for errors), it prints 12593. Which is 49*256 + 49 - in other words, "11" as an integer (ascii 1 being 49). So nothing wrong with the code as far as I can see (apart from the memory layout assumptions pointed out by Benoit), so we do need to know what you expected to happen

#include <stdio.h>
#include<ctype.h>

typedef struct {
  int Type;
  int Type2;
}foo;

typedef struct {
  char cData[40];
}bar;

int main()
{
  bar b1;
  foo *f=(foo *)&b1;
  strcpy(b1.cData,"11");
  printf("Type is  %d \n",f->Type);
  return 0;
}

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