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Pointers Pointer is a variable that contains the address of a variable Here P is sahd to point to the variable C C 7 34…… 173172174175176177178179180181.

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Presentation on theme: "Pointers Pointer is a variable that contains the address of a variable Here P is sahd to point to the variable C C 7 34…… 173172174175176177178179180181."— Presentation transcript:

1 Pointers Pointer is a variable that contains the address of a variable Here P is sahd to point to the variable C C 7 34…… …… P

2 Referencing The unary operator & gives the address of a variable The statement P=&C assigns the address of C to the variable P, and now P points to C To print a pointer, use %p format.

3 Referencing int C; int *P; /* Declare P as a pointer to int */ C = 7; P = &C; C 7 34…… …… P

4 Dereferencing The unary operator * is the dereferencing operator Applied on pointers Access the object the pointer points to The statement *P=5; Puts in C (the variable pointed by P) the value 5

5 Dereferencing printf(“%d”, *P); /* Prints out ‘7’ */ *P = 177; printf(“%d”, C); /* Prints out ‘177’ */ P = 177; /* This is unadvisable! */ C 7 34…… …… P

6 Example pointers.c

7 pointers.c – step by step int x=1, y=2, z[10]={5,6,7}; int *ip; /* ip is a pointer to int */ ip = &x; /* ip now points to x */ printf("ip now points to x that contains the value %d\n",*ip); y = *ip; /* y is now 1 */ printf("y is now %d\n",y); *ip = 0; /* x is now 0 */ printf("x is now %d\n",x); ip = &z[2]; /* ip now points to z[2] */ printf("ip now points to z[2] that contains the value %d\n",*ip); *ip = 1; /* z[2] is now 1 */ printf("z[2] is now %d\n", z[2]); printf("ip is %p\n", ip); xy zip Z[0]Z[1]Z[2] …

8 pointers.c – step by step int x=1, y=2, z[10]={5,6,7}; int *ip; /* ip is a pointer to int */ ip = &x; /* ip now points to x */ printf("ip now points to x that contains the value %d\n",*ip); y = *ip; /* y is now 1 */ printf("y is now %d\n",y); *ip = 0; /* x is now 0 */ printf("x is now %d\n",x); ip = &z[2]; /* ip now points to z[2] */ printf("ip now points to z[2] that contains the value %d\n",*ip); *ip = 1; /* z[2] is now 1 */ printf("z[2] is now %d\n", z[2]); printf("ip is %p\n", ip); xy zip Z[0]Z[1]Z[2]

9 pointers.c – step by step int x=1, y=2, z[10]={5,6,7}; int *ip; /* ip is a pointer to int */ ip = &x; /* ip now points to x */ printf("ip now points to x that contains the value %d\n",*ip); y = *ip; /* y is now 1 */ printf("y is now %d\n",y); *ip = 0; /* x is now 0 */ printf("x is now %d\n",x); ip = &z[2]; /* ip now points to z[2] */ printf("ip now points to z[2] that contains the value %d\n",*ip); *ip = 1; /* z[2] is now 1 */ printf("z[2] is now %d\n", z[2]); printf("ip is %p\n", ip); xy zip Z[0]Z[1]Z[2]

10 pointers.c – step by step int x=1, y=2, z[10]={5,6,7}; int *ip; /* ip is a pointer to int */ ip = &x; /* ip now points to x */ printf("ip now points to x that contains the value %d\n",*ip); y = *ip; /* y is now 1 */ printf("y is now %d\n",y); *ip = 0; /* x is now 0 */ printf("x is now %d\n",x); ip = &z[2]; /* ip now points to z[2] */ printf("ip now points to z[2] that contains the value %d\n",*ip); *ip = 1; /* z[2] is now 1 */ printf("z[2] is now %d\n", z[2]); printf("ip is %p\n", ip); xy zip Z[0]Z[1]Z[2]

11 pointers.c – step by step int x=1, y=2, z[10]={5,6,7}; int *ip; /* ip is a pointer to int */ ip = &x; /* ip now points to x */ printf("ip now points to x that contains the value %d\n",*ip); y = *ip; /* y is now 1 */ printf("y is now %d\n",y); *ip = 0; /* x is now 0 */ printf("x is now %d\n",x); ip = &z[2]; /* ip now points to z[2] */ printf("ip now points to z[2] that contains the value %d\n",*ip); *ip = 1; /* z[2] is now 1 */ printf("z[2] is now %d\n", z[2]); printf("ip is %p\n", ip); xy zip Z[0]Z[1]Z[2]

12 pointers.c – step by step int x=1, y=2, z[10]={5,6,7}; int *ip; /* ip is a pointer to int */ ip = &x; /* ip now points to x */ printf("ip now points to x that contains the value %d\n",*ip); y = *ip; /* y is now 1 */ printf("y is now %d\n",y); *ip = 0; /* x is now 0 */ printf("x is now %d\n",x); ip = &z[2]; /* ip now points to z[2] */ printf("ip now points to z[2] that contains the value %d\n",*ip); *ip = 1; /* z[2] is now 1 */ printf("z[2] is now %d\n", z[2]); printf("ip is %p\n", ip); xy zip Z[0]Z[1]Z[2]

13 pointers.c – step by step int x=1, y=2, z[10]={5,6,7}; int *ip; /* ip is a pointer to int */ ip = &x; /* ip now points to x */ printf("ip now points to x that contains the value %d\n",*ip); y = *ip; /* y is now 1 */ printf("y is now %d\n",y); *ip = 0; /* x is now 0 */ printf("x is now %d\n",x); ip = &z[2]; /* ip now points to z[2] */ printf("ip now points to z[2] that contains the value %d\n",*ip); *ip = 1; /* z[2] is now 1 */ printf("z[2] is now %d\n", z[2]); printf("ip is %p\n", ip); xy zip Z[0]Z[1]Z[2]

14 pointers.c – step by step int x=1, y=2, z[10]={5,6,7}; int *ip; /* ip is a pointer to int */ ip = &x; /* ip now points to x */ printf("ip now points to x that contains the value %d\n",*ip); y = *ip; /* y is now 1 */ printf("y is now %d\n",y); *ip = 0; /* x is now 0 */ printf("x is now %d\n",x); ip = &z[2]; /* ip now points to z[2] */ printf("ip now points to z[2] that contains the value %d\n",*ip); *ip = 1; /* z[2] is now 1 */ printf("z[2] is now %d\n", z[2]); printf("ip is %p\n", ip); xy zip Z[0]Z[1]Z[2]

15 pointers.c – step by step int x=1, y=2, z[10]={5,6,7}; int *ip; /* ip is a pointer to int */ ip = &x; /* ip now points to x */ printf("ip now points to x that contains the value %d\n",*ip); y = *ip; /* y is now 1 */ printf("y is now %d\n",y); *ip = 0; /* x is now 0 */ printf("x is now %d\n",x); ip = &z[2]; /* ip now points to z[2] */ printf("ip now points to z[2] that contains the value %d\n",*ip); *ip = 1; /* z[2] is now 1 */ printf("z[2] is now %d\n", z[2]); printf("ip is %p\n", ip); xy zip Z[0]Z[1]Z[2]

16 pointers.c – step by step int x=1, y=2, z[10]={5,6,7}; int *ip; /* ip is a pointer to int */ ip = &x; /* ip now points to x */ printf("ip now points to x that contains the value %d\n",*ip); y = *ip; /* y is now 1 */ printf("y is now %d\n",y); *ip = 0; /* x is now 0 */ printf("x is now %d\n",x); ip = &z[2]; /* ip now points to z[2] */ printf("ip now points to z[2] that contains the value %d\n",*ip); *ip = 1; /* z[2] is now 1 */ printf("z[2] is now %d\n", z[2]); printf("ip is %p\n", ip); xy zip Z[0]Z[1]Z[2]

17 pointers.c – step by step int x=1, y=2, z[10]={5,6,7}; int *ip; /* ip is a pointer to int */ ip = &x; /* ip now points to x */ printf("ip now points to x that contains the value %d\n",*ip); y = *ip; /* y is now 1 */ printf("y is now %d\n",y); *ip = 0; /* x is now 0 */ printf("x is now %d\n",x); ip = &z[2]; /* ip now points to z[2] */ printf("ip now points to z[2] that contains the value %d\n",*ip); *ip = 1; /* z[2] is now 1 */ printf("z[2] is now %d\n", z[2]); printf("ip is %p\n", ip); xy zip Z[0]Z[1]Z[2]

18 pointers.c – step by step int x=1, y=2, z[10]={5,6,7}; int *ip; /* ip is a pointer to int */ ip = &x; /* ip now points to x */ printf("ip now points to x that contains the value %d\n",*ip); y = *ip; /* y is now 1 */ printf("y is now %d\n",y); *ip = 0; /* x is now 0 */ printf("x is now %d\n",x); ip = &z[2]; /* ip now points to z[2] */ printf("ip now points to z[2] that contains the value %d\n",*ip); *ip = 1; /* z[2] is now 1 */ printf("z[2] is now %d\n", z[2]); printf("ip is %p\n", ip); xy zip Z[0]Z[1]Z[2]

19 Common errors It is impossible to define pointers to constants or expressions. It is also impossible to change a variable’s address (because it is not for us to determine!). Therefore, the following are errors: i = &3; j = &(k+5); k = &(a==b); &a = &b; &a = 150;

20 Pass arguments by value The functions we saw till now accepted their arguments “by value” They could manipulate the passed values They couldn’t change values in the calling function

21 Wrong Swap val_swap.c

22 How can we fix it? We can define swap so it gets pointers to integers instead of integers void swap(int *x, int *y) { …swap *x and *y… } We then call swap by swap(&x,&y); This is passing values by address

23 Right Swap add_swap.c

24 Back to scanf We can now understand the & in scanf(“%d”,&a); The argument list in scanf is simply passed by address, so scanf can change its content

25 Exercise Write a function that accepts a double parameter and returns its integer and fraction parts. Write a program that accepts a number from the user and prints out its integer and fraction parts, using this function

26 Solution dbl_split.c

27 Exercise The relation between rectangular and polar coordinates is given by – r = sqrt(x 2 +y 2 ) θ = tan -1 (y/x) Implement a function that accepts two rectangular coordinates and returns the corresponding polar coordinates Use the function atan defined in math.h

28 Solution rec_to_polar.c

29 Pointers and Arrays Recall that an array S holds the address of its first element S[0] S is actually a pointer to S[0] int S[10]; int *P; P=S; /* From now P is equivalent to S */ Both P and S are now pointing to S[0]

30 Pointer-array equivalence Arrays are actually a kind of pointers! When an array is defined, a fixed amount of memory the size of the array is allocated. The array variable is set to point to the beginning of that memory segment When a pointer is declared, it is uninitialized (like a regular variable) Unlike pointers, the value of an array variable cannot be changed

31 Passing arrays to functions Arrays can be passed to functions and have their values changed. This is possible because an array variable is actually an address. The two following function argument declarations are equivalent – int arr[] int *arr Example – vec_mul.c

32 Arrays as function arguments Functions can accept arrays as arguments Usually the array’s size also needs to be passed (why?) For example - int CalcSum(int arr[], int size); Within the function, arr is accessed in the usual way Example – calc_sum.c

33 Exercise Implement a function that accepts two integer arrays and returns 1 if they are equal, 0 otherwise Write a program that accepts two arrays of integers from the user and checks for equality

34 Solution compare_arrays.c

35 Exercise Implement the function int Subsequence(int arr1[], int size1, int arr2[], int size2); The function returns 1 iff arr2 is a subsequence of arr1 For example, if arr1 = {1, 4, 6, 8} and arr2 = {4, 6}.

36 Solution subsequence.c

37 Exercise Write a program that accepts positive integers from the user until a negative one is entered The program then displays the 5 largest integers entered Not necessarily sorted by size! Note – there’s no need to define functions other than main

38 Solution top_numbers.c


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