这是一个创建于 3092 天前的主题，其中的信息可能已经有所发展或是发生改变。
#define ARRAY_SIZE(a) (sizeof(a)/sizeof(a[0]))
int fun(void)
{
int array[] = {1, 2, 3, 4, 5};
if ((ARRAY_SIZE(array) - 2) > -1) {
/* xxx */
return 0;
}
return -1;
}
int fun(void)
{
int array[] = {1, 2, 3, 4, 5};
if ((ARRAY_SIZE(array) - 2) > -1) {
/* xxx */
return 0;
}
return -1;
}
5 条回复 • 2016-08-13 21:29:52 +08:00
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1
sen506 2016-06-12 23:28:13 +08:00 via Android  1
int 转成 uint 去运算了，，
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2
azh7138m 2016-06-12 23:31:35 +08:00 via Android
楼上正解， sizeof 返回的是无符号数，运算符两边存在无符号数的时候就会把带符号数转换为无符号数
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3
htfy96 2016-06-12 23:52:02 +08:00
因为 ARRAY_SIZE(array)-2 是 unsigned long ，比-1 的 int 阶高，所以 int 会被转成 unsigned long
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4
zhongshaohua OP 感谢楼上各位老司机
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5
liashui 2016-08-13 21:29:52 +08:00
If both operands have the same type, then no further conversion is needed.
Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer conversion rank is converted to the type of the operand with greater rank. Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand, then the operand with signed integer type is converted to the type of the operand with unsigned integer type. Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, then the operand with unsigned integer type is converted to the type of the operand with signed integer type. Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type. |
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