在C中获取浮点型变量的整型表示值 [英] get the integer representation value of a float type variable in C

问题描述

我有数字20（0x14）存储在一个32位寄存器。该寄存器被分配给一个表示值2.8e-44的C float变量。现在我想获取float变量的十六进制表示，并将其存储到一个整型变量中，以便恢复信息的原始含义。除了做一些指针业务外，还有更好的方法吗？这里是代码：

  #include< stdio.h> 
 
 int main（）
 {
 float f = 2.802597e-44; 
 int nv，* pnv; 
 printf（f =％f \ n，f）; 
 printf（n =％d \\\
，f）; 
 nv =（int）f; 
 printf（n =％d \\\
，nv）; 
 pnv =（int *）& f; 
 printf（n =％d \\\
，（* pnv））; 
返回0; 
 
 
 
 $ b我得到了我想要的pnv整数指针。有没有更好的办法做到这一点避免指针和工作在C？ 
 
解决方案
你可以实现你的需要与联合： / p> 
 
 
  #include< stdio.h> 
 
 int main（）
 {
 union {int i; float f; } fi; 
 fi.f = 2.802597e-44; 
 printf（f =％e \\\
，fi.f）; 
 printf（n =％d \\\
，fi.i）; 
返回0; 
} 
  
 
I have the number 20 (0x14) stored in a 32-bit register. The register is allocated to a C float variable representing the value 2.8e-44. Now I want to get the hexadecimal representation of the float variable back and stored it into an integer variable in order to restore the original meaning of the information. Is there a better way to do that, apart from doing some pointer business? Here is the code:
#include <stdio.h>

int main()
{
    float f = 2.802597e-44;
    int nv,*pnv;
    printf("f=%f\n",f);
    printf("n=%d\n",f);
    nv=(int)f;
    printf("n=%d\n",nv);
    pnv=(int*)&f;
    printf("n=%d\n",(*pnv));
    return 0;
}
I get what I want using the pnv integer pointer. Is there a better way to do that avoiding pointers and working in C?
解决方案
You can achieve your need with a union:
#include <stdio.h>

int main()
{
  union { int i; float f; } fi;
  fi.f = 2.802597e-44;
  printf("f=%e\n",fi.f);
  printf("n=%d\n",fi.i);
  return 0;
}


                        
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