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Ex. Linked list (資料結構會教)

Pass parameters in functions
Ex. sum
#include <stdio.h> int sum(int a, int b); void main(void) { int a=2, b=3, s; s=sum(a,b); printf("sum=%d\n", s); } int sum(int a, int b) // function can only return one value back { return (a+b); }
address
b=0xFFEA &b=0xFE09
0xFFEA 0xFFEE 0xFFF1
Question

What is wrong with this?
int *b; b=malloc(3); // 3 bytes
*b=1; // b[0]=1; *(b+1)=2; // b[1]=2; *(b+2)=3; // b[2]=3;
address
int a = 5; int *b; 指向整數變數的指標
b=an address
&b=0xFE09
a=5
&a=0xFFEA
Question about pointer

Can I assign the value of a pointer arbitrarily?

Ex.
…
nrow
s[i][j]
在函數內動態宣告陣列並回傳給 主程式使用
main() { a=???? int *a; myalloc(??); } myalloc(??) { int *b; b=malloc(3*sizeof(int)); }
address
b=0x1221
address
5 a
0xFFEA &a pointer
Pointer 指標，指位器 (cont.)

指標亦為變數，有：值、位址，指標型態


Pointer is also a variable, has: value, address, and data type 指標的值為一個位址。The value of a pointer is an address
}
4. 指標指向二維陣列

Pointer is assigned to an 2-d array
address
int a[2][3]; int *b; b=a; *b=1; // a[0][0]=1; *(b+1)=2; // a[0][1]=2; *(b+2)=3; // a[0][2]=3; *(b+3)=4; // a[1][0]=4;
a=0xFFEA
a[0] a[1] a[2]
a=0xFFEA 0xFFEE 0xFFF1
動態宣告指標陣列 dynamically allocate array of pointers

If you are not comfortable with double pointers, try this
i=0,j=ncol-1
i=nrow-1,j=0
i=1,j=0
Exercise: 動態分配記憶體做轉置 矩陣
In matrix transpose
C code
Why use pointers?

Use dynamically allocated memory High level data structure
3. 指標指向陣列

陣列名稱為一個指標常數。The name of an array variable is an constant pointer
address
int a[3]; int *b; b=a; *b=1; // b[0]=1; *(b+1)=2; // b[1]=2; *(b+2)=3; // b[2]=3;
for(i=0;i<nrow;i++) for(j=0; j<ncol; j++) *(s+i*ncol+j)=0; i=0,j=0 s i=0,j=1 s+1 …
i=1,j=0
for(i=0;i<nrow;i++) for(j=0; j<ncol; j++) *(s+i+j*nrow)=0; i=0,j=0 i=1,j=0 … s s+1
b = &a;
b=0xFFEA &b=0xFE09 a=5 &a=0xFFEA
2. 指標指向使用者動態分配記憶 體 malloc

Dynamically allocate memory for pointers
int *b;
b=malloc(3*sizeof(int));
*b=1; // b[0]=1; *(b+1)=2; // b[1]=2; *(b+2)=3; // b[2]=3;
address
#include <stdio.h> void main(void) { int *a[3];
}
printf("%p\n", a[0]);
a[0] a[1] a[2]
a=0xFFEA 0xFFEE 0xFFF1
typedef: 自訂資料型態

typedef: define a new data type #include <stdio.h> void main(void) { typedef int *INTPTR; INTPTR a[3]; } printf("%p\n", a[0]);
b=0xFFEA &b=0xFE09
a[0][0] a=0xFFEA a[0][1] 0xFFEE a[0][2] 0xFFF1 a[1][0] …
Two ways to dynamically allocate 2-d arrays (1)
char *s; s = (char *)malloc(sizeof(char)*nrow*ncol);
int * k; k=0x12; *k = 100; Ans: 指標的值必須是有意義的
1. 指標指向其他變數

利用指標指向其他變數。Use pointers to store the address of other variables.
address
int a = 5; int *b; 指向整數變數的指標
char **s; s = (char **)malloc(sizeof(char *)*nrow); for (i=0; i<nrow; i++) s[i] = (char *)malloc(sizeof(char)*ncol); s[i][j]=‘a’; /* access data */ s nrow ncol ncol
#include <stdio.h> #include <stdlib.h>
void main(void) { typedef int *INTPTR; INTPTR *a; a = malloc(3*sizeof(INTPTR)); printf("%p\n", a[0]);
}
Two ways to dynamically allocate 2-d arrays (2)
動態宣告指標陣列 dynamically allocate array of pointers
#include <stdio.h> #include <stdlib.h>
address
void main(void) { int **a; a = malloc(3*sizeof(int *));
} printf("%p\n", a[0]);
Announcement

請登入討論區網頁 依照第二次期中考及格與否，有不同作業。
Notes
請多寫程式，甚至故意把程式寫錯是很重要的Hale Waihona Puke Pointer 指標，指位器

變數有三個屬性：位址、值和資料型態 Variable has 3 properties: its address, value, and data type int a = 5;
b=0xFFEA &b=0xFE09
a[0] a[1] a[2]
a=0xFFEA 0xFFEE 0xFFF1
Question

What is wrong with this?
int a[3]; a++;
Example
#include <stdio.h> #include <stdlib.h> void main(void) { char s[]="I am an idiotI am an idiot"; int i, j; char *ptr; i=0; while(1) { ptr = s+i; for(j=0; j<13; j++) printf("%c", *(ptr+j)); i++; i = i % 14; system("cls"); }
Ex. Sum and division

How to re-write the above function to return the result of sum and division?