问题描述
可能重复：结果
1
main()
{
if(-1<(unsigned char)1)
printf("-1 is less than (unsigned char)1:ANSI semantics");
else
printf("-1 NOT less than (unsigned char)1:K&R semantics");
}
2
int array[] = {23,41,12,24,52,11};
#define TOTAL_ELEMENTS (sizeof(array)/sizeof(array[0]))
main()
{
int d = -1,x;
if(d<=TOTAL_ELEMENTS -2)
x = array[d+1];
}
第一个转换unsigned char型1到符号变量在ANSI C，
而第二个程序转换d可一个unsigned int，使
条件前pression在ANSI C.返回false
为什么他们不同的表现？
The first convert unsigned char 1 to a signed variable in ANSI C,while the second program convert d to an unsigned int that makes thecondition expression return false in ANSI C.Why did they behave differently?
推荐答案
有关第一个右手边是一个无符号的字符，以及所有无符号的字符值放入一个符号整数，所以它转换为符号整数。
For the first one the right-hand side is an unsigned char, and all unsigned char values fit into a signed int, so it is converted to signed int.
有关第二个右手边是一个unsigned int，因此左侧是由符号int转换为unsigned int类型。
For the second one the right-hand side is an unsigned int, so the left-hand side is converted from signed int to unsigned int.
又见<一个href=\"https://www.securecoding.cert.org/confluence/display/sec$c$c/INT02-C.+Understand+integer+conversion+rules\"相对=nofollow>整数转换这CERT文档。
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