Pointer
指针
- 指针是什么：指针是一个值，为内存地址的变量。
- 声明指针（
int *pi, *pi2;）：类型 * 变量名 * 变量名 - &：取地址（
ptr = &pooh;），把 pooh 的地址给 ptr。 - *：解引用（
val = *ptr;），把 ptr 所指向的数据赋值给 val。 - 运算优先级：*与++相同，顺序从右往左
优先级
“顺序从右到左”我的理解是，都在指针左边比如*++p2就先++再*，但如果是*p1++那就还是先*再++
/* order.c -- precedence in pointer operations */
#include <stdio.h>
int data[2] = {100, 200};
int moredata[2] = {300, 400};
int main(void)
{
int * p1, * p2, * p3;
p1 = p2 = data;
p3 = moredata;
printf(" *p1 = %d, *p2 = %d, *p3 = %d\n",
*p1 , *p2 , *p3);
printf("*p1++ = %d, *++p2 = %d, (*p3)++ = %d\n",
*p1++ , *++p2 , (*p3)++);
printf(" *p1 = %d, *p2 = %d, *p3 = %d\n",
*p1 , *p2 , *p3);
return 0;
}
// *p1 = 100, *p2 = 100, *p3 = 300
// *p1++ = 100, *++p2 = 200, (*p3)++ = 300
// *p1 = 200, *p2 = 200, *p3 = 301
- pintf 指针：%p
- 指向下一个内容：++
- 严重错误：解引用未初始化的指针，相当于在不知道什么地方改变了数据
解引用未初始化的指针
#include <stdio.h>
#include <stdlib.h>
int main()
{
// wrong
int *a = 1;
// wrong
int *b;
*b = 100;
// right
int *c = malloc(sizeof(int));
*c = 200;
return 0;
}
指针与数组
- p代表指针，eg
int *p; - array代表数组 eg
int array[32]; - 已经把指针指向数组：
p = array[0];
指针指向数组；通过指针访问数组
#include <stdio.h>
int main()
{
int array[8] = {0, 1, 2, 3, 4, 5, 6, 7};
int *p1 = array;
int *p2 = &array[0];
for (int i = 0; i < 8; i++)
{
printf("%d ", *(p1 + i)); // 0 1 2 3 4 5 6 7
printf("%d ", *(p2 + i)); // 0 1 2 3 4 5 6 7
}
return 0;
指针与二维数组：精髓在于，指针是“一层一层”指下去的
/* zippo1.c -- zippo info */
#include <stdio.h>
int main(void)
{
int zippo[4][2] = {{2, 4}, {6, 8}, {1, 3}, {5, 7}};
printf(" zippo = %p, zippo + 1 = %p\n",
zippo, zippo + 1);
printf("zippo[0] = %p, zippo[0] + 1 = %p\n",
zippo[0], zippo[0] + 1);
printf(" *zippo = %p, *zippo + 1 = %p\n",
*zippo, *zippo + 1);
printf("zippo[0][0] = %d\n", zippo[0][0]);
printf(" *zippo[0] = %d\n", *zippo[0]);
printf(" **zippo = %d\n", **zippo);
printf(" zippo[2][1] = %d\n", zippo[2][1]);
printf("*(*(zippo+2) + 1) = %d\n", *(*(zippo + 2) + 1));
return 0;
}
// zippo = 0x7ff7b301c7f0, zippo + 1 = 0x7ff7b301c7f8
// zippo[0] = 0x7ff7b301c7f0, zippo[0] + 1 = 0x7ff7b301c7f4
// *zippo = 0x7ff7b301c7f0, *zippo + 1 = 0x7ff7b301c7f4
// zippo[0][0] = 2
// *zippo[0] = 2
// **zippo = 2
// zippo[2][1] = 3
// *(*(zippo+2) + 1) = 3
指针与二维数组，注意，[]优先级高于*！
/* zippo2.c -- zippo info via a pointer variable */
#include <stdio.h>
int main(void)
{
int zippo[4][2] = { {2,4}, {6,8}, {1,3}, {5, 7} };
int (*pz)[2]; // pz 指向一个内部含有两个 int 的数组
pz = zippo;
printf(" pz = %p, pz + 1 = %p\n",
pz, pz + 1);
printf("pz[0] = %p, pz[0] + 1 = %p\n",
pz[0], pz[0] + 1);
printf(" *pz = %p, *pz + 1 = %p\n",
*pz, *pz + 1);
printf("pz[0][0] = %d\n", pz[0][0]);
printf(" *pz[0] = %d\n", *pz[0]);
printf(" **pz = %d\n", **pz);
printf(" pz[2][1] = %d\n", pz[2][1]);
printf("*(*(pz+2) + 1) = %d\n", *(*(pz+2) + 1));
return 0;
}
// (base) kimshan@MacBook-Pro output % ./"zippo2"
// pz = 0x7ff7bc6da800, pz + 1 = 0x7ff7bc6da808
// pz[0] = 0x7ff7bc6da800, pz[0] + 1 = 0x7ff7bc6da804
// *pz = 0x7ff7bc6da800, *pz + 1 = 0x7ff7bc6da804
// pz[0][0] = 2
// *pz[0] = 2
// **pz = 2
// pz[2][1] = 3
// *(*(pz+2) + 1) = 3
- 符合字面量：比如
(int [2]){10,20}
Demo
// flc.c -- funny-looking constants
#include <stdio.h>
#define COLS 4
int sum2d(const int ar[][COLS], int rows);
int sum(const int ar[], int n);
int main(void)
{
int total1, total2, total3;
int * pt1;
int (*pt2)[COLS];
pt1 = (int [2]) {10, 20};
pt2 = (int [2][COLS]) { {1,2,3,-9}, {4,5,6,-8} };
total1 = sum(pt1, 2);
total2 = sum2d(pt2, 2);
total3 = sum((int []){4,4,4,5,5,5}, 6);
printf("total1 = %d\n", total1);
printf("total2 = %d\n", total2);
printf("total3 = %d\n", total3);
return 0;
}
int sum(const int ar[], int n)
{
int i;
int total = 0;
for( i = 0; i < n; i++)
total += ar[i];
return total;
}
int sum2d(const int ar[][COLS], int rows)
{
int r;
int c;
int tot = 0;
for (r = 0; r < rows; r++)
for (c = 0; c < COLS; c++)
tot += ar[r][c];
return tot;
}
// (base) kimshan@MacBook-Pro output % ./"flc"
// total1 = 30
// total2 = 4
// total3 = 2
指针与函数
- 通过 int array[] 和 int *array，作为形参是一样的
int array[] 与 int *array
#include <stdio.h>
#include <stdlib.h>
void merge(int *array1, int array2[], int res[], int len1, int len2)
{
for (int i = 0; i < len1 + len2; i++)
*(res + i) = *(array1 + i);
for (int i = 0; i < +len2; i++)
res[i + len1] = array2[i];
}
int main()
{
const int LEN1 = 4;
const int LEN2 = 2;
int array1[LEN1] = {0, 1, 2, 3};
int array2[LEN2] = {4, 5};
int *array3 = malloc(sizeof(int) * (LEN1 + LEN2));
merge(array1, array2, array3, LEN1, LEN2);
for (int i = 0; i < LEN1 + LEN2; i++)
printf("%d ", *(array3 + i));
return 0;
}
- 悬空指针：指针指向一个地方，但是那个地方作用域已经结束
悬空指针（修改一行上边的代码）
#include <stdio.h>
#include <stdlib.h>
int *merge(int *array1, int array2[], int len1, int len2)
{
int res[len1 + len2];
for (int i = 0; i < len1 + len2; i++)
*(res + i) = *(array1 + i);
for (int i = 0; i < +len2; i++)
res[i + len1] = array2[i];
return res; // wrong!但是编译器能通过
}
int main()
{
const int LEN1 = 4;
const int LEN2 = 2;
int array1[LEN1] = {0, 1, 2, 3};
int array2[LEN2] = {4, 5};
int *array3 = malloc(sizeof(int) * (LEN1 + LEN2));
array3 = merge(array1, array2, LEN1, LEN2);
for (int i = 0; i < LEN1 + LEN2; i++)
printf("%d ", *(array3 + i));
return 0;
}
- 形参对原地址的变量的保护
形参+const，这样如果不小心写了array[i]++，编译器就会捕获这个错误
int sum(const int *array, int n)
{
int total = 0;
for(int i=0; i<n; i++)
total += array[i];
return total;
}
- 多重指针/多维数组，在函数里边怎么传参数
- 不能这样声明：
int sum2(int ar[][], int rows);，两个都是[]不行 - 需要这样：
int sum2(int ar[][COL], int rows);，COL 放在数组中，rows 单独传入 - 只能忽略最左边的：
int sum4d(int ar[][2][3][4], int rows);因为C 规定数组的维数必须是固定的。
- 不能这样声明：
有很多等价写法
// array2d.c -- functions for 2d arrays
#include <stdio.h>
#define ROWS 3
#define COLS 4
void sum_rows(int ar[][COLS], int rows);
void sum_cols(int[][COLS], int); // ok to omit names
int sum2d(int (*ar)[COLS], int rows); // another syntax
int main(void)
{
int junk[ROWS][COLS] = {
{2, 4, 6, 8},
{3, 5, 7, 9},
{12, 10, 8, 6}};
sum_rows(junk, ROWS);
sum_cols(junk, ROWS);
printf("Sum of all elements = %d\n", sum2d(junk, ROWS));
return 0;
}
void sum_rows(int ar[][COLS], int rows)
{
int r;
int c;
int tot;
for (r = 0; r < rows; r++)
{
tot = 0;
for (c = 0; c < COLS; c++)
tot += ar[r][c];
printf("row %d: sum = %d\n", r, tot);
}
}
void sum_cols(int ar[][COLS], int rows)
{
int r;
int c;
int tot;
for (c = 0; c < COLS; c++)
{
tot = 0;
for (r = 0; r < rows; r++)
tot += ar[r][c];
printf("col %d: sum = %d\n", c, tot);
}
}
int sum2d(int ar[][COLS], int rows)
{
int r;
int c;
int tot = 0;
for (r = 0; r < rows; r++)
for (c = 0; c < COLS; c++)
tot += ar[r][c];
return tot;
}
// (base) kimshan@MacBook-Pro output % ./"array2d"
// row 0: sum = 20
// row 1: sum = 24
// row 2: sum = 36
// col 0: sum = 17
// col 1: sum = 19
// col 2: sum = 21
// col 3: sum = 23
// Sum of all elements = 8
- 以上的写法很不方便，所以 C11 引入了 VLA 变长数组
Demo
//vararr2d.c -- functions using VLAs
#include <stdio.h>
#define ROWS 3
#define COLS 4
int sum2d(int rows, int cols, int ar[rows][cols]);
int main(void)
{
int i, j;
int rs = 3;
int cs = 10;
int junk[ROWS][COLS] = {
{2,4,6,8},
{3,5,7,9},
{12,10,8,6}
};
int morejunk[ROWS-1][COLS+2] = {
{20,30,40,50,60,70},
{5,6,7,8,9,10}
};
int varr[rs][cs]; // VLA
for (i = 0; i < rs; i++)
for (j = 0; j < cs; j++)
varr[i][j] = i * j + j;
printf("3x5 array\n");
printf("Sum of all elements = %d\n",
sum2d(ROWS, COLS, junk));
printf("2x6 array\n");
printf("Sum of all elements = %d\n",
sum2d(ROWS-1, COLS+2, morejunk));
printf("3x10 VLA\n");
printf("Sum of all elements = %d\n",
sum2d(rs, cs, varr));
return 0;
}
// function with a VLA parameter
int sum2d(int rows, int cols, int ar[rows][cols])
{
int r;
int c;
int tot = 0;
for (r = 0; r < rows; r++)
for (c = 0; c < cols; c++)
tot += ar[r][c];
return tot;
}
// (base) kimshan@MacBook-Pro output % ./"vararr2d"
// 3x5 array
// Sum of all elements = 80
// 2x6 array
// Sum of all elements = 315
// 3x10 VLA
// Sum of all elements = 270