A Note on TMin

关于表示32位整型中最小的数-2147483648的一篇文章,from CSAPP http://csapp.cs.cmu.edu/public/tmin.html

Summary

Due to one of the rules for processing integer constants in ANSI C, the numeric constant -2147483648 is handled in a peculiar way on a 32-bit, two's complement machine.

The problem can be corrected by writing -2147483647-1, rather than -2147483648 in any C code.

Description of Problem

The ANSI C standard requires that an integer constant too large to be represented as a signed integer be ``promoted'' to an unsigned value. When GCC encounters the value 2147483648, it gives a warning message: `` warning: decimal constant is so large that it is unsigned.'' The result is the same as if the value had been written 2147483648U.
ANSI C标准中要求太大的整型常量都表示为unsigned

The compiler processes an expression of the form -X by first reading the expression X and then negating it. Thus, when the C compiler encounters the constant -2147483648, it first processes 2147483648, yielding 2147483648U, and then negates it. The unsigned negation of this value is also 2147483648U. The bit pattern is correct, but the type is wrong!

书上的一个小错误

Writing TMin in Code

The ANSI C standard states that the maximum and minimum integers should be declared as constants INT_MAX and INT_MIN in the file limits.h. Looking at this file on an IA32 Linux machine (in the directory /usr/include), we find the following declarations:

/* Minimum and maximum values a `signed int' can hold.  */

#define INT_MAX	2147483647

#define INT_MIN	(-INT_MAX - 1)

This method of declaring INT_MIN avoids accidental promotion and also avoids any warning messages by the C compiler about integer overflow.

The following are ways to write TMin_32 for a 32-bit machine that give the correct value and type, and don't cause any error messages:

• -2147483647-1
• (int) 2147483648U
• 1<<31

The first method is preferred, since it indicates that the result will be a negative number.
表示-2147483648的几种方法