（浮动）铸造不起作用

Question

I', trying this simple code. It shows the first 10 integers that can not be represented in float:

int main(){
  int i, cont=0;
  float f;
  double di, df;
  for(i=10000000, f=i; i<INT_MAX; i++, f=i, df=f, di=((float)i)){
    if(i!=f){
      printf("i=%d   f=%.2f   df=%.2lf   di=%.2lf\n", i, f, df, di);
      if(cont++==10) return 0;
    }
  }
  return 1;
}

di is a double variable, but I set it to (float)i , so it should be equal to df, but it is not.

For example, the number 16777217 is represented as 16777216 by f and df , but di is still 16777217, ignoring the (float) casting.

How is this possible?

**I am using this: gcc (Ubuntu 4.4.3-4ubuntu5) 4.4.3

Answer 1

Relevant to your question is 6.3.1.8:2 in the C99 standard :

The values of floating operands and of the results of floating expressions may be represented in greater precision and range than that required by the type; the types are not changed thereby.

and in particular footnote 52:

The cast and assignment operators are still required to perform their specified conversions as described in 6.3.1.4 and 6.3.1.5.

Reading the footnote, I would say that you have identified a bug in your compiler.

You may have identified two bugs in your compiler: the i!=f comparison is done between floats (see promotion rules on the same page of the standard), so it should always be false. Although, in this latter case, I think that the compiler may be allowed to use a larger type for the comparison by 6.3.1.8:2, perhaps making the comparison equivalent to (double)i!=(double)f and thus sometimes true. Paragraph 6.3.1.8:2 is the paragraph in the standard I hate most, and I am still trying to understand strict aliasing.

Answer 2

This post explains what is going on:

http://www.exploringbinary.com/when-floats-dont-behave-like-floats/

Basically extra precision might be stored on the machine for different expression evaluations, making what would be equal floats not equal.