N.I.M : 12131291
Kelas : 12.2C.06
Soal :
Buatlah ilustrasi tabel, pemetaan RMO & CMO, Jalur perpindahan serta hitunglah hasilnya dalam Hexa array - array dibawah ini :
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Jawaban :
1. Array Float A[5] dengan nilai awal A[1] = 002F[H]. Berapa nilai A[3] ..?
Diketahui
| A[i] = A[3]
B = 2F[A] i = 3 L = 4 |
Penyelesaian
| A[i] = B + (i – 1) * L A[3] = 2F(H) + (3 – 1) * 4 A[3] = 2*17(H) + 8(D) A[3] = 34(H) + 8(H) A[3] = 42(H) |
0 1 2 3 4
| A[1] | A[2] | A[3] | A[4] | A[5] |
2. Array Float A[10] dengan nilai awal A[8] = 012F[H]. Berapa nilai A[3]..?
Diketahui
| A[i] = A[3]
B = 12F[A] i = 3 L = 4 |
Penyelesaian
| A[i] = B + (i – 1) * L A[3] = 12F(H) + (3 – 1) * 4 A[3] = 12*17(H) + 8(D) A[3] = 204(H) + 8(H) A[3] = 212(H) |
0 1 2 3 4 5 6 7 8 9
| A[1] | A[2] | A[3] | A[4] | A[5] | A[6] | A[7] | A[8] | A[9] | A[10] |
3. Array Float A[5][4] dengan nilai awal A[1][0] = 002F[H]. Berapa nilai A[3][3]?
| Index | 0 | 1 | 2 | 3 |
| 0 | ||||
| 1 | 2F(H) | |||
| 2 | ||||
| 3 | ? | |||
| 4 |
Rumus
| RMO : @M[i][j] = M[0][0] + {(j - 1) * K + (i - 1)} * L
CMO : @M[i][j] = M[0][0] + {(i - 1) * N + (j - 1)} * L |
Keterangan
| @M[i][j] = Posisi Array yg dicari M[0][0] = Posisi alamat awal index Array i = Baris j = kolom L = Ukuran memory type data K = Banyaknya elemen per kolom N = Banyaknya elemen per baris |
Jawaban Secara Baris Per Baris (Row Major Oder / RMO)
Jalur Perpindahan
Jalur Perpindahan
| Float 4 byte Perpindahan baris : 2 - 0 = 2 * 4 (kolomnya) = 8 Perpindahan kolom : 3 - 0 = 3 Total perpindahan : 8 + 3 = 11 |
| A[1][0] >> A[1][1] >> A[1][2] >> A[1][3] >> A[2][0] >> A[2][1] >> A[2][2] >> A[2][3] >> A[3][0] >> A[3][1] >> A[3][2] >> A[3][3] |
Hasil
| @A[i][j] = @A[1][0] + {(i - 1) * N + (j - 1)} * L A[3][3] = 2F(H) + {(5 - 1) * 5 + (4 - 1)} * 4 A[3][3] = 2*17(H) + 23 * 4 A[3][3] = 34(H) + 92(D) A[3][3] = 34(H) + 5C(H) A[3][3] = 34(H) + 60(H) A[3][3] = 94(H) |
Jawaban Secara Kolom Per Kolom (Coloumn Major Oder / CMO)
Jalur Perpindahan
Jalur Perpindahan
| Float 4 byte Perpindahan baris : 3 - 0 = 2 * 5 (kolomnya) = 15 Perpindahan kolom : 2 - 0 = 2 Total perpindahan : 15 + 2 = 17 |
| A[2][0] >> A[3][0] >> A[4][0] >> A[0][1] >> A[1][1] >> A[2][1] >> A[3][1] >> A[4][1] >> A[0][2] >> A[1][2] >> A[2][2] >> A[3][2] >> A[4][2] >> A[0][3] >> A[1][3] >> A[2][3] >> A[3][3] |
Hasil
| @M[i][j] = @M[1][0] + {(j - 1) * K + (i - 1)} * L A[3][3] = 2F(H) + {(4 - 1) * 4 + (5 - 1)} * 4 A[3][3] = 2*17(H) + 16 *4 A[3][3] = 34(H) + 64(D) A[3][3] = 34(H) + 40(H) A[3][3] = 74(H) |
4. Array Int A[5][4] dengan nilai awal A[3][3] = 00CF[H]. Berapa nilai A[0][1]?
| Index | 0 | 1 | 2 | 3 |
| 0 | ? | |||
| 1 | ||||
| 2 | ||||
| 3 | C(F) | |||
| 4 |
Rumus
| RMO : @M[i][j] = M[0][0] + {(j - 1) * K + (i - 1)} * L
CMO : @M[i][j] = M[0][0] + {(i - 1) * N + (j - 1)} * L |
Keterangan
| @M[i][j] = Posisi Array yg dicari M[0][0] = Posisi alamat awal index Array i = Baris j = Kolom L = Ukuran memory type data K = Banyaknya elemen per kolom N = Banyaknya elemen per baris |
Jawaban Secara Baris Per Baris (Row Major Oder / RMO)
Jalur Perpindahan
Jalur Perpindahan
| Float 4 byte Int 2 byte Perpindahan baris : 0 - 3 = -3 * 4 (kolomnya) = -12 Perpindahan kolom : 1 - 3 = -2 Total perpindahan : -12+ (-2) = -14 |
| A[3][2] >> A[3][1] >> A[3][0] >> A[2][3] >> A[2][2] >> A[2][1] >> A[2][0] >> A[1][3] >> A[1][2] >> A[1][1] >> A[1][0] >> A[0][3] >> A[0][2] >> A[0][1] |
Hasil
| @A[i][j] = @A[3][3] + {(i - 1) * N + (j - 1)} * L A[0][1] = CF(H) - {(5 - 1) * 5 + (4 - 1)} * 2 A[0][1] = 12*17(H) - 23 * 2 A[0][1] = 204(H) - 46(D) A[0][1] = 204(H) - 2E(H) A[0][1] = 204(H) - 32(H) A[0][1] = 172(H) |
Jawaban Secara Kolom Per Kolom (Coloumn Major Oder / CMO)
Jalur Perpindahan
Jalur Perpindahan
| Int 2 byte Perpindahan baris : 1 - 3 = -2 * 5 (kolomnya) = -10 Perpindahan kolom : 0 - 3 = -3 Total perpindahan : -10 + (-3) = -13 |
| A[2][3] >> A[1][3] >> A[0][3] >> A[4][2] >> A[3][2] >> A[2][2] >> A[1][2] >> A[0][2] >> A[4][1] >> A[3][1] >> A[2][1] >> A[1][1] >> A[0][1] |
Hasil
| @M[i][j] = @M[3][3] - {(j - 1) * K + (i - 1)} * L A[0][1] = CF(H) - {(4 - 1) * 4 + (5 - 1)} * 2 A[0][1] = 12*17(H) + 16 * 4 A[0][1] = 204(H) - 64(D) A[0][1] = 204(H) - 40(H) A[0][1] = 164(H) |
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