426C 传递数组给函数

Posted 2020-11-10 alex_bn_lee

参考：C 传递数组给函数

参考：C语言二维数组作为函数参数传递

参考：二维数组作为函数参数传递剖析(C语言)(6.19更新第5种)

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总结：

一维数组参数，可以是地址、arr[]、arr[n]

二维数组比较复杂，需要提供列的值，否则会报错，可参考上面的网址

#include <stdio.h>
#include <stdlib.h>

void myFunc1(int length, int *arr) {
	for (int i = 0; i < length; i++) {
		printf("%d ", arr[i]);
	}
	printf("\\n");
	return;
}

void myFunc2(int length, int arr[]) {
	for (int i = 0; i < length; i++) {
		printf("%d ", arr[i]);
	}
	printf("\\n");
	return;
}

void myFunc3(int length, int arr[length]) {
	for (int i = 0; i < length; i++) {
		printf("%d ", arr[i]);
	}
	printf("\\n");
	return;
}

void myFunc4(int row, int column, int *arr2) {
	for (int i = 0; i < row; i++) {
		for (int j = 0; j < column; j++) {
			printf("arr2[%d][%d] = %d   ", i, j, *(arr2 + i*column + j));
		}
		printf("\\n");
	}
	return;
}

void myFunc5(int row, int column, int arr2[][column]) {
	for (int i = 0; i < row; i++) {
		for (int j = 0; j < column; j++) {
			printf("arr2[%d][%d] = %d   ", i, j, arr2[i][j]);
		}
		printf("\\n");
	}
	return;
}

int main() {
	int arr[5] = {1, 2, 3, 4, 5};
	
	printf("Parameter is pointer:\\n");
	myFunc1(5, &arr[0]);	// ok with arr
	
	printf("Parameter is array[]:\\n");
	myFunc2(5, &arr[0]);	// ok with arr
	
	printf("Parameter is array[length]:\\n");
	myFunc3(5, &arr[0]);	// ok with arr
	
	int arr2[2][3] = {{1, 2, 3}, {4, 5, 6}};
	
	printf("Parameter is pointer:\\n");
	myFunc4(2, 3, arr2[0]);	// &arr2[0][0] is ok, but arr2 is not ok
							// *arr2 is ok.
							// but actually they have the same address
	
	printf("Parameter is pointer:\\n");
	myFunc5(2, 3, arr2);	// not ok with arr2[0]
							// not ok with &arr2[0][0]
	
	printf("\\narr2                           = %p\\n", arr2);
	printf("arr2[0]                        = %p\\n", arr2[0]);
	printf("*arr2 = arr2[0]                = %p\\n", *arr2);
	printf("arr2[0][0]                     = %p\\n", &arr2[0][0]);
	printf("**arr2 = *arr2[0] = arr2[0][0] = %p\\n\\n", &arr2[0][0]);
	
	
	printf("address of this two-dimentional array:\\n");
	printf("arr2           = %p\\n", arr2);
	for (int i = 0; i < 2; i++) {
		printf("  arr2[%d]      = %p\\n", i, arr2[0]);
		for (int j = 0; j < 3; j++) {
			printf("    arr2[%d][%d] = %p   ", i, j, &arr2[i][j]);
		}
		printf("\\n");
	}
	
	printf("we can use the unexisted element, interesting!\\n");
	printf("*arr2 means arr2[0], the first row.\\n");
	printf("**arr2 means *arr2[0] means arr2[0][0], the first element of the first row.\\n");
	printf("arr2[1][0]               = %d\\n", arr2[1][0]);
	printf("arr2[0][3]               = %d\\n", arr2[0][3]);
	printf("*(*(arr2 + 1) + 0)       = %d\\n", *(*(arr2 + 1) + 0));
	printf("*(*(arr2 + 0) + 3)       = %d\\n", *(*(arr2 + 0) + 3));
	printf("*(arr2[0] + 1*3 + 0)     = %d\\n", *(arr2[0] + 1*3 + 0));
	printf("*(*arr2 + 1*3 + 0)       = %d\\n", *(*arr2 + 1*3 + 0));
	printf("*(&arr2[0][0] + 1*3 + 0) = %d\\n", *(&arr2[0][0] + 1*3 + 0));
	printf("*(&arr2[0][0] + 3)       = %d\\n", *(&arr2[0][0] + 3));
	
	return 0;
}

output：

Parameter is pointer:
1 2 3 4 5 
Parameter is array[]:
1 2 3 4 5 
Parameter is array[length]:
1 2 3 4 5 
Parameter is pointer:
arr2[0][0] = 1   arr2[0][1] = 2   arr2[0][2] = 3   
arr2[1][0] = 4   arr2[1][1] = 5   arr2[1][2] = 6   
Parameter is pointer:
arr2[0][0] = 1   arr2[0][1] = 2   arr2[0][2] = 3   
arr2[1][0] = 4   arr2[1][1] = 5   arr2[1][2] = 6   

arr2                           = 0x7ffc6bdc38a0
arr2[0]                        = 0x7ffc6bdc38a0
*arr2 = arr2[0]                = 0x7ffc6bdc38a0
arr2[0][0]                     = 0x7ffc6bdc38a0
**arr2 = *arr2[0] = arr2[0][0] = 0x7ffc6bdc38a0

address of this two-dimentional array:
arr2           = 0x7ffc6bdc38a0
  arr2[0]      = 0x7ffc6bdc38a0
    arr2[0][0] = 0x7ffc6bdc38a0       arr2[0][1] = 0x7ffc6bdc38a4       arr2[0][2] = 0x7ffc6bdc38a8   
  arr2[1]      = 0x7ffc6bdc38a0
    arr2[1][0] = 0x7ffc6bdc38ac       arr2[1][1] = 0x7ffc6bdc38b0       arr2[1][2] = 0x7ffc6bdc38b4   
we can use the unexisted element, interesting!
*arr2 means arr2[0], the first row.
**arr2 means *arr2[0] means arr2[0][0], the first element of the first row.
arr2[1][0]               = 4
arr2[0][3]               = 4
*(*(arr2 + 1) + 0)       = 4
*(*(arr2 + 0) + 3)       = 4
*(arr2[0] + 1*3 + 0)     = 4
*(*arr2 + 1*3 + 0)       = 4
*(&arr2[0][0] + 1*3 + 0) = 4
*(&arr2[0][0] + 3)       = 4