#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 转成 uint 去运算了，，

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

因为 ARRAY_SIZE(array)-2 是 unsigned long ，比-1 的 int 阶高，所以 int 会被转成 unsigned long

感谢楼上各位老司机

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.