CODEWITHSUNDEEP
cfloating-point
judoka_acl
judoka_acl

Reputation: 385

get the integer representation value of a float type variable in C

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?

Upvotes: 2

Views: 645

Answers (6)

mouviciel
mouviciel

Reputation: 67929

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;
}

Upvotes: 5

Bathsheba
Bathsheba

Reputation: 234885

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++.

Upvotes: 2

Aboudi
Aboudi

Reputation: 318

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?

Upvotes: -2

user
user

Reputation: 143

you can use type casting ..

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

Upvotes: 1

gsamaras
gsamaras

Reputation: 73444

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?

Upvotes: 0

zgrw
zgrw

Reputation: 127

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

Upvotes: 1

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