当铸造指针的增量操作者会发生什么？ [英] What happens when a casted pointer has an increment operator?

问题描述

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);

编译上面与++运算符的行生成一个左值所需的错误。

Compiling the above generates an lvalue required error on the line with the ++ operator.

推荐答案

演员创建了一个类型的临时指针为int * 。可以不增加临时因为它不表示存储结果的地方。

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 ++ standardese，（INT *）P 是右值，大致意思是前pression，只有能发生在赋值的右手边。

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 ，另一方面是一个左值，这意味着它可以有效地出现在赋值的左侧。只有左值可以增加。

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.

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