Thread: Type conversion from float to double

Hi to all,

I'm studying C from "C Primer Plus" ... and it says that on page 174:

"When appearing in an expression, char and short, both signed and unsigned, are automatically converted to int or, if necessary, to unsigned int. (If short is the same size as int, unsigned short is larger than int; in that case, unsigned short is converted to unsigned int.) Under K&R C, but not under current C, float is automatically converted to double. Because they are conversions to larger types, they are called promotions."

In my compiler (Visual Studio), float constants are automatically converted to doubles. So, do you think that the sentence about the conversion from float to double correct?

Thank you Elkvis.

Originally Posted by Salem

If you use the 'f' suffix, then your float constants really are float constants.
Without a suffix, the constants inherently have a type 'double'.

So, you as well think that the underlined sentence from the book is probably incorrect ... thanks a lot.

float constants are NOT automatically converted to doubles. That would be idiotic. What would be the point of having float constants then? If you just mean something like 1.23, then that's a double already. 1.23f is a float constant.

For example, I see no evidence of conversion to double in the assembly code for the following (rather contrived) code.

Code:

#include <stdio.h>

int main() {
    int n = 1;
    n *= 1.23f;
    printf("%d\n", n);
    return 0;
}

The assembly stores the 1.23f value as

Code:

    .align 4
.LC0:
    .long    1067282596

which is the 4-byte bit pattern for 1.23f.

Changing 1.23f to 1.23 yields

Code:

    .align 8
.LC0:
    .long    2061584302
    .long    1072934420

which is the 8-byte bit pattern for 1.23.

However, passing a float to a variadic function (as one of the variadic arguments) will convert it to double.

Code:

short sh1 = 10;
short sh2 = 11;
short sh3 = sh1 + sh2;

In this code, as far as I know, sh1 and sh2 are converted to int before adding them and then the resulting int value is trimmed to short when assigning to sh3, right?

What about the same example for floating points:

Code:

float fl1 = 10.0;
float fl2 = 11.0;
float fl3 = fl1 + fl2;

I think that the underlined section in the quoted paragraph refers to this case. Here, are fl1 and fl2 converted to double before adding them?