Lecture 5: Multidimensional Arrays. Wednesday, 11 February 2009

Transcription

1 Lecture 5: Multidimensional Arrays CS209 : Algorithms and Scientific Computing Wednesday, 11 February 2009 CS209 Lecture 5: Multidimensional Arrays 1/20

2 In today lecture... 1 Let s recall... 2 Multidimensional Arrays 3 The sizeof() things 4 Exercise 5 Dynamic memory allocation A silly joke An example CS209 Lecture 5: Multidimensional Arrays 2/20

3 Overview Last week... Pointers Arrays This week: Multidimensional arrays : e.g., matrices Dynamic Memory allocation: how to request space for an array on-the-fly. Strings: to to use arrays of characters to store words and lines of text. Further details can be found in A Book on C, Chapter 6. CS209 Lecture 5: Multidimensional Arrays 3/20

4 Let s recall... An array is used to store the an ordered, indexed list of items, all of the same data type. To declare an an array/vector of integers a 0, a 1,..., a 9 we use int a[10]; An array can be assigned values during declaration: char s[6]={ h, e, l, l, o, \0 }; When an array is declared, e.g., int a[10]; the system declares a pointer to int called a. the pointer is to a[0] space is allocated to the addresses at a, a+1,... The address stored in a cannot be changed. CS209 Lecture 5: Multidimensional Arrays 4/20

5 Multidimensional Arrays If an array (particularly of integers or floats) is like a mathematical vector, then how do we define a matrix? A matrix is a two-dimensional array. For example, to declare a 3 3 matrix of floats, we would use the syntax: float A[3][3]; So A[0][0] A[0][1] A[0][2] A = A[1][0] A[1][1] A[1][2] A[2][0] A[2][1] A[2][2] In general an n m array is declared as float A[n][m]; CS209 Lecture 5: Multidimensional Arrays 5/20

6 Multidimensional Arrays If a program has the line: int a[3][2]; What really happens is that the system creates 3 arrays, each of length 2. More precisely, it declares 3 pointers to type int: a[0], a[1], and a[2], space for storing an integer is allocated to each of the address a[0], a[0]+1, and a[1], a[1]+1, a[2], and a[2]+1. CS209 Lecture 5: Multidimensional Arrays 6/20

7 Multidimensional Arrays This means that if a[][] is declared as a two-dimensional 3 4 array, then the following are equivalent: a[1][2] *(a[1] + 2) *( *(a + 1) + 2) *( &a[0][0] ) CS209 Lecture 5: Multidimensional Arrays 7/20

8 Multidimensional Arrays 01Matrix.c /* An example of a program with a 2d array */ #include <stdio.h> int main(void ) { int a[3][4]={1,2,3,4,5,6,7,8,9,10,11,12}; } printf("a[1][2] = %d\n", a[1][2]); printf("*(a[1]+2) = %d\n", *(a[1] + 2)); printf("*(*(a+1)+2) = %d\n", *( *(a + 1) + 2)); printf("*(&a[0][0] ) = %d\n", *( &a[0][0] )); return(0); CS209 Lecture 5: Multidimensional Arrays 8/20

9 Multidimensional Arrays In another example, we ll sum all the entries of a 3 4 array. Example (02Sum a Matrix.c) /* Summing the elements of a 2d array */ #include <stdio.h> int sum(int a[][4]); int main(void ) { int n; int a[3][4]={1,2,3,4,5,6,7,8,9,10,11,12}; } n = sum(a); printf("sum of the entries of a: %d \n",n); return(0); CS209 Lecture 5: Multidimensional Arrays 9/20

10 Multidimensional Arrays Example int sum(int a[][4]) { int i,j, ans=0; for (i=0; i < 3; i++) for (j=0; j< 4; j++) ans += a[i][j]; } return(ans); Important: Notice that this function is defined only for arrays of size 3 4. Even if we passed n and m as arguments to the function, we would still have to declare that a has 4 columns. CS209 Lecture 5: Multidimensional Arrays 10/20

11 The sizeof() things Before studying a better way of handling arrays, we work out how much storage is required by integers, floats, characters, and even pointers. This is because we need to tell the system how many bytes are required to store an array. Two problems with this might be: we have enough things to remember without knowing how many bytes in a float. some systems store data types differently from others. Later we ll use complex structures which have variable memory requirements. CS209 Lecture 5: Multidimensional Arrays 11/20

12 The sizeof() things The good news is that the sizeof() operator returns the number of bytes for a particular data type. It can take data types (e.g, float, char) or variable names as its argument. It returns an unsigned integer (Why?). SizeOf.c int x=-123, *p; char name[6]="cs209"; printf("a char takes\t %lu bytes;\n", sizeof(char)); printf("a float uses\t %lu bytes;\n", sizeof(float)); printf("but a double uses %lu bytes;\n", sizeof(double)); printf("x is requires\t%l3u bytes;\n", sizeof(x)); printf("a pointer needs\t%l3u bytes;\n", sizeof(p)); printf("array %s is stored in %l3u bytes.\n", name, sizeof(name)); CS209 Lecture 5: Multidimensional Arrays 12/20

13 Exercise Write a short C program to check how much memory is required to store a pointer to type int a pointer to a pointer to type int (i.e, int **) a pointer to type double, float, char CS209 Lecture 5: Multidimensional Arrays 13/20

14 Dynamic memory allocation Having to declare the size of an array in a function header is a huge restriction. Either we have to modify the program every time we change the size of the array, or we have to define the array to be as big as possible. This is wasteful of time and resources. To minimise the amount of coding we have to do, and to use resources well, we simply declare an appropriate variable with type pointer to int for an integer vector: int *v pointer to pointer to int for an integer matrix: int **m CS209 Lecture 5: Multidimensional Arrays 14/20

15 Dynamic memory allocation Next we must ask the system to reserve some memory. There are four important commands: sizeof() (see above) calloc(n, sizeof(x)): Continuous Memory Allocation. It will reserve enough space to n variables each with same size as x. It sets than all to zero. malloc(n*sizeof(int)): Memory Allocation. As above, but it doesn t do any initialization. free(ptr): deallocate the space the begins at the address stored in ptr. The headers for these functions is in stdlib.h. CS209 Lecture 5: Multidimensional Arrays 15/20

16 Dynamic memory allocation The important thing is that we can call calloc(), malloc() and free() any time we like. This is what is dynamic about it. Both malloc() and calloc() return pointers to the base of the memory they allocated. However, because they are not for specific pointer types (e.g., pointer to int or pointer to char) they return void pointers. These void pointers are then re-cast. E.g., c = (char *) malloc(7*sizeof(char)); d = (float *) calloc(10, sizeof(float)); CS209 Lecture 5: Multidimensional Arrays 16/20

17 Dynamic memory allocation A silly joke Lost in translation Fingering the ultimate nothingness that underlies everything. Translation of pointer to void found in a Japanese programming manual. CS209 Lecture 5: Multidimensional Arrays 17/20

18 Dynamic memory allocation An example In the example below, the size of the vector v is not fixed until the program is run. Note that the sum function will work for any sized array. CS209 Lecture 5: Multidimensional Arrays 18/20

19 Dynamic memory allocation An example Dynamic.c #include <stdlib.h> int sum(int *v, int n); int main(void ) { int *v, n, i, ans; printf("how many elements are there in v? :"); scanf("%d", &n); v=(int *) calloc(n, sizeof(int)); for (i=0; i < n; i++) { printf("enter v[%d]: ", i); scanf("%d", &v[i]); } CS209 Lecture 5: Multidimensional Arrays 19/20

20 Dynamic memory allocation An example } ans= sum(v, n); printf("the sum of the entries is %d \n", ans); free(v); return(0); int sum(int *u, int n) { int i, ans=0; for (i=0; i < n; i++) ans += u[i]; } return(ans); CS209 Lecture 5: Multidimensional Arrays 20/20