当强制转换的指针具有增量运算符时会发生什么?

What happens when a casted pointer has an increment operator?(当强制转换的指针具有增量运算符时会发生什么?)

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.

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