IEEE Representation of Floats

It is my understanding that float in c is stored as IEEE single precision representation. However

void main (int argv, char argc[])
{
float a = -341.274;
printf("%X\n",a);
return;
}

When I execute that, the result is:

C0755462

The IEEE 754 single precision representation for -341.274 is: C3AAA312
and double precision is: C07554624DD2F1AA

So wtf is happening here? It seems like the float is stored as a double precision but then for some reason truncated after 8 bits. Whats going on!??!?

Yes, printf receives all float values as "double" - this is part of the C specification for "variadic arguments" (that is, functions that take variable number of arguments). And it's not truncated, it's your output that is truncated.

If you want to print the hex representation of a number, you could do something like this:

Code:

---

  printf("%X", *(unsigned int *)&x);

---

That forces the compiler to generate code that interprets the float value as an integer for passing to printf, instead of promoting it to double.

--
Mats

1. Your main() is all wrong.
2. Use [code][/code] tags around your code.

Quote:

---

Originally Posted by matsp

Yes, printf receives all float values as "double" - this is part of the C specification for "variadic arguments" (that is, functions that take variable number of arguments). And it's not truncated, it's your output that is truncated.

If you want to print the hex representation of a number, you could do something like this:

Code:

---

  printf("%X", *(unsigned int *)&x);

---

That forces the compiler to generate code that interprets the float value as an integer for passing to printf, instead of promoting it to double.

--
Mats

---

Thanks a lot! I understand it really well and my program works perfectly, absolutely fantastic.

Quote:

---

Originally Posted by Pnevma

It is my understanding that float in c is stored as IEEE single precision representation.

---

Your understanding is incorrect.