类型为int的指针指向浮点指针 [英] Typecasting int pointer to float pointer

问题描述

  int a [8] = {1,2,3,4， 5,6,7,8}; 
 
 printf（％f\\\
，*（float *）a）; 
 printf（％f\\\
，*（（float *）a + 1））; 
 printf（％f\\\
，*（（float *）a + 2））; 
 printf（％f\\\
，*（（float *）a + 3））; 
 printf（％f \ n，*（（float *）a + 4））; 
 printf（％f\\\
，*（（float *）a + 5））; 
 printf（％f\\\
，*（（float *）a + 6））; 
 printf（％f\\\
，*（（float *）a + 7））; 
  

我得到

  0.000000 
 0.000000 
 0.000000 
 0.000000 
 0.000000 
 0.000000 
 0.000000 
 0.000000 
  float * 的函数的参数传递。  
 
 
看起来这样做效果不好。有人可以解释为什么这是行不通的？ 
  int * ptr; 
函数（（float *）ptr）; 
  
如果我这样做，函数不读取指针正确指向的值。  0.0000 。
解决方案
这是不正确的。  int 和 float 不保证具有相同的对齐方式。  
 
 
记住：投射值和投射指针是不同的情况。投射指针会改变方式来引用类型值，几乎肯定会导致中的大部分 / b> 
 
 
根据 C11 标准文档，第6.3.2.3节 
 
 
 
 
指向对象类型的指针可能被转换为指向不同对象类型的指针。如果生成的指针没有正确对齐引用类型，那么行为是不确定的。
 
在你的情况下，一个解决方法可能是 
 
 
  printf（％f \ n ，（float）* a）; //转换值，而不是指针
  
 
I'm trying to do the following
int a[8]={1,2,3,4,5,6,7,8};

printf("%f\n", *(float *)a);
printf("%f\n", *((float *)a+1));
printf("%f\n", *((float *)a+2));
printf("%f\n", *((float *)a+3));
printf("%f\n", *((float *)a+4));
printf("%f\n", *((float *)a+5));
printf("%f\n", *((float *)a+6));
printf("%f\n", *((float *)a+7));
I get 
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
The reason why I'm trying to print the elements in this way is because, I want to cast the int pointer to the array to the float pointer and pass it as a parameter for another function which only takes float *. 


It seems that this does not work well. Can someone explain why this is not working? 
int *ptr;
function((float *)ptr);
If I do this the function does not read the values the pointer is pointing to properly.. just returning 0.0000.
解决方案
This is not correct. int and float are not guaranteed to have the same alignment. 

Remember: Casting a value and casting a pointer are different scenarios. Casting a pointer changes the way to refer to the type value, which can almost certainly result in a mis-alignment in most of the cases.

As per C11 standard document, chapter §6.3.2.3

  
A pointer to an object type may be converted to a pointer to a different object type. If the resulting pointer is not correctly aligned68) for the referenced type, the behavior is undefined.
In your case, a work-around may be
printf("%f\n", (float)*a);  //cast the value, not the pointer


                        
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