Computer Programming for Engineering Applica4ons. Intro to Programming 10/22/13 1 ECE 175. Mul4- dimensional Arrays

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1 Computer Programming for Engineering Applica4ons ECE 75 Intro to Programming Mul4- dimensional Arrays Declara4on of arrays with more than one dimension Syntax: data type array_name[size][size] Example: int two_dimensions[5][7]; " double many_dimensions[0][20][30][50]; Accessing arrays with mul4ple dimensions Example: int two_dimensions[5][7]; "x = two_dimensions[][4]; "y = two_dimensions[2][4] + two_dimensions[3][]; 2 Subscripts of Two Dimensional Arrays j: i: Example: int two_dimensions[7][5]; "x = two_dimensions[][4]; // x =? "y = two_dimensions[2][4] + two_dimensions[0][]; "// y =? 3

2 Ini4alizing arrays with more than one dimension Nested for loops Declare two indexes that would run through the size of each dimension Write two for nested for loops Example: "int two_dimensions[5][7]; " "int, i, j; " "for (i=0; i<5; i++) " "{ " " "for (j=0; j<7; j++) " " " two_dimensions[i][j]=0; " 4 Ini4alizing Mul4- dimensional Arrays at Declara4on You can ini4alize mul4- dimensional arrays during declara4on Syntax: data type array_name[size][size]={ {v, v2, v3..vn, {x, x2, x3...xn,...,{z, z2,... zn; Example: int scores[3][4]={ {0,0,0,0,{0,0,0,0, " {0,0,0,0 ; float cells[][2]={ {0.0,, {.5, 2, {0.3, 2, {0, 0 You can omit only the first dimension of the array 5 Accessing a subset of table elements Use if condi4ons on the indexes to the array elements Example: "int table[4][4]; " "int, i,j; " "for (i=0; i<4; i++) " "{ " " for (j=0; j<4; j++) { " " if (i >= j) " " table[i][j]=(i+)*(j+); " "

3 Passing mul4dimensional arrays to func4ons Syntax: data type fun_name(int ar_name[][size], int var Examples: void addition(int x[][5], int y[][5], int rows) void threed(float z[][6][0]) Only the first dimension can be omited, when arrays are passed As in the case of one dimension, a pointer to the first element of the array is passed to the func4on There is a way to use pointers, but it is somewhat confusing. S4ck to this nota4on 7 Adding two matrices Write a C program that adds two two- dimensional matrices and prints out the result Program should have two func4ons, one that facilitates adding matrices, and one that facilitates prin4ng. 8 Body of main #include<stdio.h> #define ROW 5 #define COL 5 void print_array(int x[][col]); void add_array(int x[][col], int y[][col], int z[][col]); " int main(void) { " int mat[row][col]={{,,,,; " int mat2[row][col]; " int result[row][col]; " int i,j; " for (i=0; i<row; i++) { for (j=0; j<col; j++) { mat[i][j]=i+j; mat2[i][j]=(i+)*(j+); " add_array(mat,mat2,result); print_array(mat); printf(" \n"); print_array(mat2); printf(" \n"); print_array(result); return(0); 9 3

4 Func4on for adding two 2- dim arrays void add_array(int x[][col], int y[][col], int z[][col]) { int i, j; " for (i=0; i<row; i++) { for (j=0; j<col; j++) " z[i][j]=x[i][j]+y[i][j]; 0 Func4on for prin4ng 2- dim array Prints each row, then print a new line. void print_array(int x[][col]) { int i,j; " for (i=0; i<row; i++) { "for (j=0; j<col; j++) " " "{ printf("%4d",x[i][j]); " printf("\n"); Game Fun Let s Play Tic- Tac- Toe Write a program that checks who won on a 4c- tac- toe game O X O O X X X X O 2 4

5 Body of main #include<stdio.h> #define P 0 #define P2 #define TIE 2 int check_fun(int x[][3]); // Checking the winner of a tic-tac-toe board int main(void){ int i,j; // indexes for accessing rows and columns int board[3][3]; // the tic-tac-toe board FILE *inp=fopen("game.dat", "r"); if (inp==null) // if file does not exist printf("unable to load game file"); else{ for(i=0; i<3; i++){ for(j=0; j<3; j++) fscanf(inp, "%d", &board[i][j]); // storing everyting on board if (check_fun(board)==p) // calling check_fun function printf("the Xs Won\n"); else if (check_fun(board)==p2) printf("the Os Won\n"); else printf("this is a Tie\n"); return 0; 3 The Checking Func4on int check_fun(int x[][3]) { int i; int outcome; // returned value for (i=0; i<3; i++) { // row sum or column sum or diagonal sum is equal to 3 if (x[i][0]+x[i][]+x[i][2]==3 x[0][i]+x[][i]+x[2] [i]==3 x[0][0]+x[][]+x[2][2]==3 x[0][2]+x[][]+x[2][0]==3) " { outcome = P; break; else if (x[i][0]+x[i][]+x[i][2]==0 x[0][i]+x[][i]+x[2] [i]==0 x[0][0]+x[][]+x[2][2]==0 x[0][2]+x[][]+x[2][0]==0) { outcome = P2; break; else outcome=tie; return outcome; 4 5