当强制转换的指针具有增量运算符时会发生什么?
例如:
int x[100];
void *p;
x[0] = 0x12345678;
x[1] = 0xfacecafe;
x[3] = 0xdeadbeef;
p = x;
((int *) p) ++ ;
printf("The value = 0x%08x", *(int*)p);
编译上述代码会在带有 ++ 运算符的行上生成一个 lvalue required 错误.
Compiling the above generates an lvalue required error on the line with the ++ operator.
推荐答案
强制转换创建了一个 int *
类型的临时指针.您不能增加临时值,因为它不表示存储结果的位置.
The cast creates a temporary pointer of type int *
. You can't increment a temporary as it doesn't denote a place to store the result.
在 C 和 C++ 标准中,(int *)p
是一个 rvalue,大??致表示只能出现在赋值右侧的表达式.
In C and C++ standardese, (int *)p
is an rvalue, which roughly means an expression that can only occur on the right-hand side of an assignment.
p
是一个左值,这意味着它可以有效地出现在赋值的左侧.只有左值可以递增.
p
on the other hand is an lvalue, which means it can validly appear on the left-hand side of an assignment. Only lvalues can be incremented.
相关文章