1

I have to change this code of 2D array to 1D array but i am not sure how to do it. Because the values of the 2D arrays are in hexadecimal.

int Sbox[SIZE_A][SIZE_B] = { //two dimensional array
{
    0x63, 0x7c, 0x77, 0x7b, 0xf2, 0x6b, 0x6f, 0xc5, 0x30, 0x01, 0x67, 0x2b,
    0xfe, 0xd7, 0xab, 0x76
}, //*  initializers for row indexed by 0 
{
    0xca, 0x82, 0xc9, 0x7d, 0xfa, 0x59, 0x47, 0xf0, 0xad, 0xd4, 0xa2, 0xaf,
    0x9c, 0xa4, 0x72, 0xc0
}, //*  initializers for row indexed by 1
{
    0xb7, 0xfd, 0x93, 0x26, 0x36, 0x3f, 0xf7, 0xcc, 0x34, 0xa5, 0xe5, 0xf1,
    0x71, 0xd8, 0x31, 0x15
}, //*  initializers for row indexed by 2
{
    0x04, 0xc7, 0x23, 0xc3, 0x18, 0x96, 0x05, 0x9a, 0x07, 0x12, 0x80, 0xe2,
    0xeb, 0x27, 0xb2, 0x75
}, {
    0x09, 0x83, 0x2c, 0x1a, 0x1b, 0x6e, 0x5a, 0xa0, 0x52, 0x3b, 0xd6, 0xb3,
    0x29, 0xe3, 0x2f, 0x84
}, {
    0x53, 0xd1, 0x00, 0xed, 0x20, 0xfc, 0xb1, 0x5b, 0x6a, 0xcb, 0xbe, 0x39,
    0x4a, 0x4c, 0x58, 0xcf
}, {
    0xd0, 0xef, 0xaa, 0xfb, 0x43, 0x4d, 0x33, 0x85, 0x45, 0xf9, 0x02, 0x7f,
    0x50, 0x3c, 0x9f, 0xa8
}, {
    0x51, 0xa3, 0x40, 0x8f, 0x92, 0x9d, 0x38, 0xf5, 0xbc, 0xb6, 0xda, 0x21,
    0x10, 0xff, 0xf3, 0xd2
}, {
    0xcd, 0x0c, 0x13, 0xec, 0x5f, 0x97, 0x44, 0x17, 0xc4, 0xa7, 0x7e, 0x3d,
    0x64, 0x5d, 0x19, 0x73
}, {
    0x60, 0x81, 0x4f, 0xdc, 0x22, 0x2a, 0x90, 0x88, 0x46, 0xee, 0xb8, 0x14,
    0xde, 0x5e, 0x0b, 0xdb
}, {
    0xe0, 0x32, 0x3a, 0x0a, 0x49, 0x06, 0x24, 0x5c, 0xc2, 0xd3, 0xac, 0x62,
    0x91, 0x95, 0xe4, 0x79
}, {
    0xe7, 0xc8, 0x37, 0x6d, 0x8d, 0xd5, 0x4e, 0xa9, 0x6c, 0x56, 0xf4, 0xea,
    0x65, 0x7a, 0xae, 0x08
}, {
    0xba, 0x78, 0x25, 0x2e, 0x1c, 0xa6, 0xb4, 0xc6, 0xe8, 0xdd, 0x74, 0x1f,
    0x4b, 0xbd, 0x8b, 0x8a
}, {
    0x70, 0x3e, 0xb5, 0x66, 0x48, 0x03, 0xf6, 0x0e, 0x61, 0x35, 0x57, 0xb9,
    0x86, 0xc1, 0x1d, 0x9e
}, {
    0xe1, 0xf8, 0x98, 0x11, 0x69, 0xd9, 0x8e, 0x94, 0x9b, 0x1e, 0x87, 0xe9,
    0xce, 0x55, 0x28, 0xdf
}, {
    0x8c, 0xa1, 0x89, 0x0d, 0xbf, 0xe6, 0x42, 0x68, 0x41, 0x99, 0x2d, 0x0f,
    0xb0, 0x54, 0xbb, 0x16
}
};

I want to parse the exact values 0x63, 0x7c, 0x77, etc into the 1D array. So I thought of changing the int to string first, then read out the string as 0x63, 0x7c, 0x77, etc to the 1D array. Any help on how to do it?

4

6 回答 6

2

您首先要记住的是，十六进制只是表示整数值的一种方式，它们不会以十六进制形式存储在计算机中。

第二件事是，如果您想要一个由矩阵组成的单个数组，那么该单个数组的大小将是（在您的情况下）SIZE_A * SIZE_B条目很大。

那么这只是一个遍历矩阵并设置数组中的值的问题。

于 2013-11-12T13:30:52.167 回答
1

您可以使用以下代码（行优先和列优先顺序变体 - 请参阅代码中的注释）：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define A_SIZE 2
#define B_SIZE 3

int main() {
  int values_2d[A_SIZE][B_SIZE] = {{0x12, 0x24, 0x46},{0x13,0x35, 0x57}};
  int values_1d[A_SIZE*B_SIZE]  = {0};
  int i,j;

  for (i = 0; i < A_SIZE; i++) {
    for (j = 0; j < B_SIZE; j++) {
      printf("v[%d,%d]=0x%02X\n", i, j, values_2d[i][j]);
    }
  }
  for (i = 0; i < A_SIZE*B_SIZE; i++) {
    values_1d[i] = values_2d[i/(B_SIZE)][i%B_SIZE]; // <<-- row-major order
    //values_1d[i] = values_2d[i%A_SIZE][i/A_SIZE]; // <<-- column-major order

    printf("values_1d[%d] = %d (0x%02X), values_2d[%d][%d] = %d (0x%02x)\n", 
        i, values_1d[i], values_1d[i], 
        i/B_SIZE, i%B_SIZE, 
        values_2d[i/B_SIZE][i%B_SIZE], values_2d[i/B_SIZE][i%B_SIZE]);
  }
  return 0;
}

结果：

┌─(55:55:55)─(michael@lorry)─(~/tmp/hex)
└─► gcc -o main main.c; ./main
values_1d[0] = 18 (0x12), values_2d[0][0] = 18 (0x12)
values_1d[1] = 36 (0x24), values_2d[0][1] = 36 (0x24)
values_1d[2] = 70 (0x46), values_2d[0][2] = 70 (0x46)
values_1d[3] = 19 (0x13), values_2d[1][0] = 19 (0x13)
values_1d[4] = 53 (0x35), values_2d[1][1] = 53 (0x35)
values_1d[5] = 87 (0x57), values_2d[1][2] = 87 (0x57)

它是按行优先顺序排列的。

于 2013-11-12T13:39:21.830 回答
1

这些是我的代码。可以吗？

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>

#define SIZE_A 16 //width
#define SIZE_B 16 //heightint Sbox[SIZE_A][SIZE_B] = {     //two dimensional array
{0x63, 0x7c, 0x77, 0x7b, 0xf2, 0x6b, 0x6f, 0xc5, 0x30, 0x01, 0x67, 0x2b,
0xfe, 0xd7, 0xab, 0x76}, //*  initializers for row indexed by 0 
{0xca, 0x82, 0xc9, 0x7d, 0xfa, 0x59, 0x47, 0xf0, 0xad, 0xd4, 0xa2, 0xaf,
0x9c, 0xa4, 0x72, 0xc0}, //*  initializers for row indexed by 1
{0xb7, 0xfd, 0x93, 0x26, 0x36, 0x3f, 0xf7, 0xcc, 0x34, 0xa5, 0xe5, 0xf1,
0x71, 0xd8, 0x31, 0x15}, //*  initializers for row indexed by 2
{0x04, 0xc7, 0x23, 0xc3, 0x18, 0x96, 0x05, 0x9a, 0x07, 0x12, 0x80, 0xe2,
0xeb, 0x27, 0xb2, 0x75},
{0x09, 0x83, 0x2c, 0x1a, 0x1b, 0x6e, 0x5a, 0xa0, 0x52, 0x3b, 0xd6, 0xb3,
0x29, 0xe3, 0x2f, 0x84},
{0x53, 0xd1, 0x00, 0xed, 0x20, 0xfc, 0xb1, 0x5b, 0x6a, 0xcb, 0xbe, 0x39,
0x4a, 0x4c, 0x58, 0xcf},
{0xd0, 0xef, 0xaa, 0xfb, 0x43, 0x4d, 0x33, 0x85, 0x45, 0xf9, 0x02, 0x7f,
0x50, 0x3c, 0x9f, 0xa8},
{0x51, 0xa3, 0x40, 0x8f, 0x92, 0x9d, 0x38, 0xf5, 0xbc, 0xb6, 0xda, 0x21,
0x10, 0xff, 0xf3, 0xd2},
{0xcd, 0x0c, 0x13, 0xec, 0x5f, 0x97, 0x44, 0x17, 0xc4, 0xa7, 0x7e, 0x3d,
0x64, 0x5d, 0x19, 0x73},
{0x60, 0x81, 0x4f, 0xdc, 0x22, 0x2a, 0x90, 0x88, 0x46, 0xee, 0xb8, 0x14,
0xde, 0x5e, 0x0b, 0xdb},
{0xe0, 0x32, 0x3a, 0x0a, 0x49, 0x06, 0x24, 0x5c, 0xc2, 0xd3, 0xac, 0x62,
0x91, 0x95, 0xe4, 0x79},
{0xe7, 0xc8, 0x37, 0x6d, 0x8d, 0xd5, 0x4e, 0xa9, 0x6c, 0x56, 0xf4, 0xea,
0x65, 0x7a, 0xae, 0x08},
{0xba, 0x78, 0x25, 0x2e, 0x1c, 0xa6, 0xb4, 0xc6, 0xe8, 0xdd, 0x74, 0x1f,
0x4b, 0xbd, 0x8b, 0x8a},
{0x70, 0x3e, 0xb5, 0x66, 0x48, 0x03, 0xf6, 0x0e, 0x61, 0x35, 0x57, 0xb9,
0x86, 0xc1, 0x1d, 0x9e},
{0xe1, 0xf8, 0x98, 0x11, 0x69, 0xd9, 0x8e, 0x94, 0x9b, 0x1e, 0x87, 0xe9,
0xce, 0x55, 0x28, 0xdf},
{0x8c, 0xa1, 0x89, 0x0d, 0xbf, 0xe6, 0x42, 0x68, 0x41, 0x99, 0x2d, 0x0f,
0xb0, 0x54, 0xbb, 0x16}
};

int myarray[SIZE_A*SIZE_B]={0};

void print2darray(int Sbox[SIZE_A][SIZE_B])
{
int i,j;

for (i = 0; i < SIZE_A; i++ )
{
    for (j = 0; j < SIZE_B; j++ )
    {
        printf("SBox[%d][%d] = 0x%02X\n",i,j,Sbox[i][j]);
        }
    }

for (i = 0; i < SIZE_A; i++ )
{
    for (j = 0; j < SIZE_B; j++ )
    {
        myarray[i*SIZE_A+j]=Sbox[i][j];
        int value=myarray[i*SIZE_B+j];
        printf("myarray is [%d,%d] = 0x%02X\n",i,j,value);
        }
    }

    return;
}

int main()
{
    print2darray(Sbox);
    return 0;
}
于 2013-11-14T02:18:47.447 回答
0

您可以声明一个指向int并分配内存并进行转换的指针，如图所示...

int i, j, k=0;
int *ptr = malloc(sizeof(*ptr) * SIZE_A * SIZE_B);

for(i = 0; i< SIZE_A; i++)
    for(j = 0; j< SIZE_B; j++)
    {
        ptr[k++] = Sbox[i][j];
    }
于 2013-11-12T13:33:17.050 回答
0

这应该可以解决您的问题，如果您想更改尺寸，请切换外循环和内循环

int array1D[SIZEA * SIZEB];
int index1D = 0;

for(int i = 0; i < SIZEB; ++i)
{
  for(int j = 0; i < SIZEA; ++j)
  {
    array1D[index1D++] = Sbox[j][i];
  }
}

Nishith 更快

于 2013-11-12T13:33:20.927 回答
0

关于： 我不知道该怎么做。因为二维数组的值 是十六进制的。

int Sbox[SIZE_A][SIZE_B] = {...

int Sbox 是您的第一个线索。好消息，无需转换。

每个块中表示的数字只是整数，以十六进制格式显示。因此，不需要转换（即从十六进制到整数），因为它们已经 int是. 您所要做的就是选择一种方法（在另一个答案中描述）来重新排列值。

于 2013-11-12T16:17:16.503 回答