数组指针如何存储其大小? [英] How does an array pointer store its size?

问题描述

#include "stdio.h"

#define COUNT(a) (sizeof(a) / sizeof(*(a)))

void test(int b[]) {
  printf("2, count:%d\n", COUNT(b));
}

int main(void) {
  int a[] = { 1,2,3 };

  printf("1, count:%d\n", COUNT(a));
  test(a);

  return 0;
}

结果很明显:

1, count:3
2, count:1

我的问题:

1. 声明"a"时存储的长度(计数/大小)信息在哪里?
2. 当将"a"传递给test()函数时，为什么长度(计数/尺寸)信息丢失了?

推荐答案

在C语言中没有数组指针"之类的东西.

There's no such thing as "array pointer" in C language.

大小未存储在任何地方. a不是指针，a是类型为int[3]的对象，这是编译器在编译时众所周知的事实.因此，当您要求编译器在编译时计算sizeof(a) / sizeof(*a)时，编译器就会知道答案是3.

The size is not stored anywhere. a is not a pointer, a is an object of type int[3], which is a fact well known to the compiler at compile time. So, when you ask the compiler to calculate sizeof(a) / sizeof(*a) at compile time the compiler knows that the answer is 3.

当您将a传递给函数时，您有意要求编译器将数组类型转换为指针类型(因为您已将函数参数声明为指针).对于指针，您的sizeof表达式产生完全不同的结果.

When you pass your a to the function you are intentionally asking the compiler to convert array type to pointer type (since you declared the function parameter as a pointer). For pointers your sizeof expression produces a completely different result.

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