C - 指针大小 [英] C - SizeOf Pointers

问题描述

char c[] = {'a','b','c'};
int* p = &c[0];
printf("%i\n", sizeof(*p)); //Prints out 4
printf("%i\n", sizeof(*c)); //Prints out 1

我对这部分代码感到非常困惑.p 和 c 都表示数组 c 在第 0 个索引处的地址.但是为什么 sizeof(*p) 会打印出 4?不应该是1吗?

I am extremely confused about this section of code. Both p and c represent the address of the array c at the 0th index. But why does sizeof(*p) print out 4? Shouldn't it be 1?

推荐答案

因为 p 是 int * 类型，所以 *p 是int 类型，在您的实现中显然是 4 个字节宽.

Because p is of type int *, so *p is of type int, which is apparently 4 bytes wide on your implementation.

如果您不希望程序调用未定义的行为，请使用 %zu 打印 size_t(sizeof 产生的结果).

And use %zu for printing size_t (what sizeof yields) if you don't want your program to invoke undefined behavior.

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