array [1] [2]和array [1,2]有什么区别? [英] what is the difference between array[1][2] and array[1,2]?

问题描述

如果我的二维数组是int array [4] [5] = {1,2,3,4,5,6,7,8,9,0,11,12,13,14,15,16 ，17,18,19,20};

if my two dimensional array is int array[4][5]= {1,2,3,4,5,6,7,8,9,0,11,12,13,14,15,16,17,18,19,20};

当我打印位置时:printf(％d \ n"，array [2,0]);.这是在不提供&的情况下打印值的地址.但是如果我这样打印:printf(％d \ n"，& array [2] [0]);它是值的打印地址.正如我给出的&它打印地址.但两者的地址不同.但是如果我打印printf(，:％d \ n"，array [3,0]);它正在打印数组[2,0]的相同地址.

when I print a position: printf("%d\n",array[2,0]);. this is printing the address of the value without giving the &. but if I print like this: printf("%d\n",&array[2][0]); it is printing address of the value. As I have given the & it prints the address. but both the address are different. but if I print printf(",:%d\n",array[3,0]); it is printing the same address of array[2,0].

[2] [0]和[2,0]访问数组元素之间的区别是什么.

what is the difference between [2][0] and [2,0] accessing the array elements.

推荐答案

阅读有关逗号运算符的信息. array[2][0]正在访问第三行第一列的元素. array[2,0]首先执行操作(2,0)(返回0)，因此等效于array[0]，后者返回数组的第一行(实际上是&array[0][0]，因为它会衰减为指针).

Read about comma operator. array[2][0] is accessing element at third row, first column. array[2,0] is first executing operation (2,0) (which returns 0) and thus is equivalent to array[0], which returns the first row of array (actually &array[0][0] because it decays to a pointer).

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