1 Pointers ( מצביעים ). 2 Variables in memory Primitives Arrays

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Pointers Pointers ((מצביעיםמצביעים))

2

Variables in memory

Primitives Arrays

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Pointers

Pointer is a variable that contains the address of a variable

Here P is said to point to the variable C

C

7 3 4… …

173172 174 175 176 177 178 179 180 181

174 3 4… …

P

833832 834 835 836 837 838 839 840 841

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Brief Summary of today’s session…

&x – address (pointer) of variable x

*x – content in address x (common) usage: int x = *y; (common) usage: printf(“%d”,*y);

int */double */char * - define pointer to the corresponding primitive

(common) usage: int * x = &y; int ** ?

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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.

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Referencing

int C;int *P; /* Declare P as a pointer to int

*/C = 7;P = &C;

C

7 3 4… …

173172 174 175 176 177 178 179 180 181

174 3 4… …

P

833832 834 835 836 837 838 839 840 841

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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

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Dereferencing

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

C

7 3 4… …

173172 174 175 176 177 178 179 180 181

174 3 4… …

P

833832 834 835 836 837 838 839 840 841

177

177

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Example

pointers.c

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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);

x y

z ip

1 2

364

5 6 7

Z[0] Z[1] Z[2]

120 248

364 368 372

…

564 772

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x y

z ip

1 2

364

5 6 7

Z[0] Z[1] Z[2]

120 248

364 368 372

120

564 772

12




x y

z ip

1 2

364

5 6 7

Z[0] Z[1] Z[2]

120 248

364 368 372

120

564 772

13




x y

z ip

1 1

364

5 6 7

Z[0] Z[1] Z[2]

120 248

364 368 372

120

564 772

14




x y

z ip

1 1

364

5 6 7

Z[0] Z[1] Z[2]

120 248

364 368 372

120

564 772

15




x y

z ip

0 1

364

5 6 7

Z[0] Z[1] Z[2]

120 248

364 368 372

120

564 772

16




x y

z ip

0 1

364

5 6 7

Z[0] Z[1] Z[2]

120 248

364 368 372

120

564 772

17




x y

z ip

0 1

364

5 6 7

Z[0] Z[1] Z[2]

120 248

364 368 372

372

564 772

18




x y

z ip

0 1

364

5 6 7

Z[0] Z[1] Z[2]

120 248

364 368 372

372

564 772

19




x y

z ip

0 1

364

5 6 1

Z[0] Z[1] Z[2]

120 248

364 368 372

372

564 772

20




x y

z ip

0 1

364

5 6 1

Z[0] Z[1] Z[2]

120 248

364 368 372

372

564 772

21




x y

z ip

0 1

364

5 6 1

Z[0] Z[1] Z[2]

120 248

364 368 372

372

564 772

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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;

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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

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Wrong swap (val_swap.c)

void swap(int x, int y) {int temp;

temp=x;x=y;y=temp;

}

int main(void) { int x=1,y=2;

printf("before swap: x=%d, y=%d\n",x,y);swap(x,y);printf("after swap: x=%d, y=%d\n",x,y);

}

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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

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Right Swap (add_swap.c)

void swap(int *x, int *y) { int temp;

temp=*x; *x=*y; *y=temp;}

int main(void) { int x=1,y=2;

printf("before swap: x=%d, y=%d\n",x,y); swap(&x,&y); printf("after swap: x=%d, y=%d\n",x,y);}

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Insights 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

Other relevant examples from the past?

Can we now “return” more then a single value from a function? How?

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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

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Solution

dbl_split.c

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Exercise @ home

The relation between rectangular and polar coordinates is given by –

r = sqrt(x2+y2)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

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Solution

rec_to_polar.c

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Exercise

Implement a function that accepts an integer array, and finds the two numbers that are closest together

For example, if the array is –{1, 5, 7, 10, 6, 19}

The function should find 5 and 6 (or 6 and 7, it doesn’t matter)

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Solution

array.c

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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]

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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

Documents

1 Pointers ( מצביעים ). 2 Variables in memory Primitives Arrays