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C: Casting minimum 32-bit integer (-2147483648) to float gives positive number (2147483648.0)

I was working on an embedded project when I ran into something which I thought was strange behaviour. I managed to reproduce it on codepad (see below) to confirm, but don't have any other C compilers on my machine to try it on them.

Scenario: I have a #define for the most negative value a 32-bit integer can hold, and then I try to use this to compare with a floating point value as shown below:

#define INT32_MIN (-2147483648L)

void main()
{
    float myNumber = 0.0f;
    if(myNumber > INT32_MIN)
    {
        printf("Everything is OK");
    }
    else
    {
        printf("The universe is broken!!");
    }
}

Codepad link: http://codepad.org/cBneMZL5

To me it looks as though this this code should work fine, but to my surprise it prints out The universe is broken!!.

This code implicitly casts the INT32_MIN to a float, but it turns out that this results in a floating point value of 2147483648.0 (positive!), even though the floating point type is perfectly capable of representing -2147483648.0.

Does anyone have any insights into the cause of this behaviour?

CODE SOLUTION: As Steve Jessop mentioned in his answer, limits.h and stdint.h contain correct (working) int range defines already, so I'm now using these instead of my own #define

PROBLEM/SOLUTION EXPLANATION SUMMARY: Given the answers and discussions, I think this is a good summary of what's going on (note: still read the answers/comments because they provide a more detailed explanation):

  • I'm using a C89 compiler with 32-bit longs, so any values greater than LONG_MAX and less or equal to ULONG_MAX followed by the L postfix have a type of unsigned long
  • (-2147483648L) is actually a unary - on an unsigned long (see previous point) value: -(2147483648L). This negation operation 'wraps' the value around to be the unsigned long value of 2147483648 (because 32-bit unsigned longs have the range 0 - 4294967295).
  • This unsigned long number looks like the expected negative int value when it gets printed as an int or passed to a function because it is first getting cast to an int, which is wrapping this out-of-range 2147483648 around to -2147483648 (because 32-bit ints have the range -2147483648 to 2147483647)
  • The cast to float, however, is using the actual unsigned long value 2147483648 for conversion, resulting in the floating-point value of 2147483648.0.
like image 248
avanzal Avatar asked Jul 18 '12 07:07

avanzal


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1 Answers

Replace

#define INT32_MIN (-2147483648L)

with

#define INT32_MIN (-2147483647 - 1)

-2147483648 is interpreted by the compiler to be the negation of 2147483648, which causes overflow on an int. So you should write (-2147483647 - 1) instead.
This is all C89 standard though. See Steve Jessop's answer for C99.
Also long is typically 32 bits on 32-bit machines, and 64 bits on 64-bit machines. int here gets the things done.

like image 155
WiSaGaN Avatar answered Sep 17 '22 16:09

WiSaGaN