converting data types

Is it possible for a float type variable to be placed in an area of memory set up as double when using the scanf() function?

My line of thinking was that a float is 4 bytes and a double is 8 bytes, so this conversion should be ok, it’s like 1/2 a litre into 1 litre right?

I am confused over a question posed to me that states that this conversion is not possible; maybe I do not understand the question properly. The question is:

"When utilising the scanf() function, why is it not possible for a float type variable to be placed in an area of memory set up as double?"

Thanks,

Use "%f" and a pointer to a float to scan a float
Use "%Lf" and a pointer to a double to scan a double.
So long as the conversion and the pointer agree, there shouldn't be a problem

Quote:

---

Originally Posted by Salem

Use "%f" and a pointer to a float to scan a float
Use "%Lf" and a pointer to a double to scan a double.

---

You meant "%lf"

I am sure it has something to do with how scanf itself functions and nothing to do with whether or not the data types can actually be converted. Maybe somebody who has the C standard laying around would care to enlighten us all.

BTW Salem I believe he is asking something like this:

Code:

---


double mydouble;

scanf("%f", &mydouble);

---

[edit]

Now I am not up to speed on binary representation of floating point numbers but I would have to guess that scanf does not perform implicit conversions if the type specifier and the variable pointer are of 2 different types. Thus even though a float can be promoted to a double scanf simply dumps the float (real*4) into the high order bits of the double pointer(real*8), thus you will always wind up with a really large negative number.

Again I could be wrong here as I do not know the interworkings of scanf so do not take this as absolute truth. Just a theory because I cannot seem to find any exact description at this time.

[/edit]

>Maybe somebody who has the C standard laying around would care to enlighten us all.
It's a type mismatch, what more can be said? Because scanf is a variadic function, it can't police what types you give it and has to assume that you did what you said you would do in the format string.

Quote:

---

Originally Posted by andyhunter

Code:

---

double mydouble;

scanf("%f", &mydouble);

---

---

According to the standard, this clearly invokes an undefined behaviour.

Should be

Code:

---

float mydouble;

scanf("%f", &mydouble);

---

or

Code:

---

double mydouble;

scanf("%lf", &mydouble);

---

>> I would have to guess that scanf does not perform implicit conversions if the type specifier and the variable pointer are of 2 different types.

correct, scanf() has no way of knowing what type of data is being passed to it (nor any other variable-argument processing function for that matter) - it relies solely on the format string to obtain that information.

Alright, now why don't we take this thread a little of topic adn perhaps make a decent reference thread.

I know there are people who are currently attempting to code an OS / make a server / are comfortable with assembly language around here. Additionally I know that there are quite a few comp sci majors around here. So I pose this question, could somebody explain the binary represenation of floating point numbers and exactly why the above example results in 'undefined behavior'.

PS - just pretend I am a 3 year old when explaining.

And if nobody wants to do that maybe at least some good references could be provided.

Thanks for resolving my issue Sebastiani and andyhunter,

Just to clarify then. So because scanf cannot look at the type of data that it is reading in, it would not be possible for any converting of data types to take place, as you said, it "does not perform implicit conversions"

That would be it.

Quote:

---

Originally Posted by andyhunter

So I pose this question, could somebody explain the binary represenation of floating point numbers and exactly why the above example results in 'undefined behavior'.

---

A floating point it generally is composed of

• One sign bit
• Some bits for the mantissa
• Some bits for the exponent (one of them being the sign of it)

The internal representation of the floating point numbers in C is not defined by the standard, but by the implementation. Details are given in <float.h> (range, field witdh, etc.)

The scanf() finction is 'variadic'. It means that its prototype has an ellipse (...) as last parameter.

This kind of function is tricky, and the expected type depends on the format parameter.

Without an explicit cast, a variadic function performs some implicit type conversion (called 'promotion'), so that

• a char is extended to an int
• a float is extended to a double

The addresses (all the scanf() variadic parameters are addresses) remain unchanged, but the function has no idea about the pointed type. This is indicated by the format.

Here are some examples with scanf():

• "%d" expects that the string is representing a decimal signed value, and that the destination is the address of an int
  
• "%s" expects that the string is representing a sequence of characters (until the next 'blank'), and that the destination is the address of an array of char.
  
• "%f" expects that the string is representing a floating point value, and that the destination is the address of a float.
  
• "%lf" expects that the string is representing a floating point value, and that the destination is the address of a double.

Meaning that if you say: "this is the address of a double", the conversion will be 'in double', fine, but the resulting value will be copied to the destination, using sizeof (double) bytes (say, 6 or 8).

Of course, if the destination is a float (sizeof (float) is generally 4), it happens an overflow by 2 or 4 bytes, hence the undefined behaviour (UB). (Anything can happen).

Additionally, even if the sizes were compatible, the coding will be different hence another reason for a UB.

Maybe my sig makes more sense to you know...