关于 C 的 arithmetic conversion (进行 算术运算 时的 强制转换规则)

那本书里面都有啊啊!!

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2015-3-17更新~~~~~~~~~~~~~~~~~~~

先上两个解释我的疑惑的链接:
http://en.cppreference.com/w/cpp/language/operator_arithmetic https://msdn.microsoft.com/en-us/library/3t4w2bkb.aspx
开始我是看 <<Expert C programming -- Deep C Secrets>>这本书(中文译作 C专家编程), chapter 1 里面的how quite is a quite change 这一小节, 有这样一段代码:

#include <stdio.h>

int main()

{

if(-1 < (unsigned char)1)

printf("-1 is less than (unsigned char)1: ANSI semantics.\n");

else

printf("-1 is NOT less than (unsigned char)1: K&R semantics.\n");

return 0;
}

我用了vs2013和gcc 4.9.1分别去编译运行, 都是ANSI的语义.打印第一条语句.
然后改成这样

#include <stdio.h>

int main()

{

if(-1 < (unsigned int)1)//或者是unsigned

printf("1111111111.\n");

else

printf("222222222222222.\n");

return 0;

}

gcc 4.9.1编译运行(未加任何特殊编译选项)的结果是打印第二条. 而vs2013默认编译不通过, error:负数转变成了无符号数.

开始看 <<Expert C programming -- Deep C Secrets>>这本书这里时有点偷懒, 只记得了这两句话:

Operands with different types get converted when you do arithmetic. Everything is converted to the type of the floatest, longest operand, signed if possible without losing bits.

gcc警告强度开大一点就好了

sh-4.3# gcc -o main *.c -Wall
sh-4.3# gcc -o main *.c -Wall -Wextra
main.c: In function 'main':
main.c:5:11: warning: comparison between signed and unsigned integer expressions [-Wsign-com
pare]
if(-1 < (unsigned ) 1) ^

实际上完整的规则还是本文开头的哪两个链接靠谱.

我觉得 best practice应当是尽量少用强制转换, 谁想去记忆那些无聊的规则.