获取C中float类型变量的整数表示值

Question

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?

Answer 1

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;
}

Answer 2

Note that the behaviour of (int*)&f is undefined as the pointer types are unrelated. So don't approach the problem in that way.

Having checked that sizeof(float) is the same as sizeof(int) , you could do this in one of two ways:

1) Type pruning through a union consisting of a float , and an int . Set the union using one member, and read it back with the other.

2) memcpy the contents of a variable of one type to the location of the variable of the other type.

Of these I prefer (2): (1) might be undefined with older C standards, and (2) also works well with C++.

Answer 3

You can directly cast it to integer as;

float a = 7.4;
int b = a; // this will be rounded to 7 and you will lose information

Or you can use some built-int functions like round, ceil, floor etc.

For reference: http://www.cplusplus.com/reference/cmath/round/?kw=round

Answer 4

you can use type casting ..

float x =3.4;
int y = (int)x;

Answer 5

What are you doing there is Undefined Behavior , didn't you check the warning?

warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘double’ [-Wformat=]
     printf("n=%d\n",f);
     ^

Read this please: How do the digits 1101004800 correspond with the number 20?

Answer 6

C is considered a weakley typed langauge, which may allow to assign values that belong to different types than the variable they are being assigned with, therefore you can simply do this:

 int integer = 1;  
 float floater =1.1111;

 floater = integer;  

This is known as Implicit type conversion, also known as coercion, is an automatic type conversion by the compiler. Some programming languages allow compilers to provide coercion; others require it.

but consider the following:

• what happens when the float is less than zero?