问题
int val = 5;
printf("%d",++val++); //gives compilation error : '++' needs l-value
int *p = &val;
printf("%d",++*p++); //no error
Could someone explain these 2 cases? Thanks.
回答1:
++val++ is the same as ++(val++). Since the result of val++ is not an lvalue, this is illegal. And as Stephen Canon pointed out, if the result of val++ were an lvalue, ++(val++) would be undefined behavior as there is no sequence point between the ++s.
++*p++ is the same as ++(*(p++)). Since the result of *(p++) is an lvalue, this is legal.
回答2:
The expression ++val++ is the same as (++val)++ (or perhaps ++(val++), anyway it's not very relevant). The result of the ++ operator is not the variable, but the value, and you can't apply the operator to a value.
The expression ++*p++ is the same as ++(*(p++)). The result of p++ is the value, but the result of *(p++) is a memory location, which the ++ operator can be applied to.
回答3:
also note that you're changing the address of the pointer by
int k = ++*p++;
回答4:
int j = ++val++; //gives compilation error
That because you cannot pre-increment an rvalue. ++val++ is interpreted as ++(val++) because post-increment operator has higher precedence than pre-increment operator. val++ returns an rvalue and pre-increment operator requires its operand to be an lvalue. :)
int k = ++*p++; //no error
++*p++ is interpreted as ++(*(p++)), which is perfectly valid.
来源：https://stackoverflow.com/questions/3660048/explanation-of-val-and-p-in-c