Posted on 2007-09-19 21:26 ZelluX
阅读(483) 评论(1) 编辑 收藏
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
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
.'' The result is the same as if the value had been
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!
Writing TMin in Code
The ANSI C standard states that the maximum and minimum integers
should be declared as constants INT_MAX
in the file limits.h
. Looking at
this file on an IA32 Linux machine (in the directory
), 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
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:
- (int) 2147483648U
The first method is preferred, since it indicates that the
result will be a negative number.