Question

假设我在C中有一个数组

int array[6] = {1,2,3,4,5,6}

我怎么能把它分成

{1,2,3}

和

{4,5,6}

这可以使用memcpy吗？

谢谢，

NONONO

Answer 1

不确定。直接的解决方案是使用malloc分配两个新数组，然后使用memcpy将数据复制到两个数组中。

int array[6] = {1,2,3,4,5,6}
int *firstHalf = malloc(3 * sizeof(int));
if (!firstHalf) {
  /* handle error */
}

int *secondHalf = malloc(3 * sizeof(int));
if (!secondHalf) {
  /* handle error */
}

memcpy(firstHalf, array, 3 * sizeof(int));
memcpy(secondHalf, array + 3, 3 * sizeof(int));

但是，如果原始数组存在的时间足够长，您可能甚至不需要这样做。您可以通过使用指向原始数组的指针将数组“拆分”为两个新数组：

int array[6] = {1,2,3,4,5,6}
int *firstHalf = array;
int *secondHalf = array + 3;

Answer 2

// create space for 6 ints and initialize the first 6
int array[6] = {1,2,3,4,5,6};
// reserve space for two lots of 3 contiguous integers
int one[3], two[3]; 
// copy memory of the first 3 ints of array to one
memcpy(one, array, 3 * sizeof(int)); 
// copy 3 ints worth of memory from the 4th item in array onwards
memcpy(two, &array[3], 3 * sizeof(int));

Answer 3

您不必拆分它们。如果你有

int *b = array + 3;

你有第二个阵列。将数组传递给函数时，无论如何它都变成了指针。

Answer 4

查看memcpy的工作原理，无需专门拆分数组。目标数组中所做的更改将自动转到源数组，反之亦然。

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

double *** double3d(long int dim1,long int dim2,long int dim3)
{
    long int i,j,k;
    double ***array;
    array=(double ***)malloc(dim1*sizeof(double **));
    for(i=0;i<dim1;i++)
    {
     array[i]=(double **)malloc(dim2*sizeof(double *));
     for(j=0;j<dim2;j++)
      array[i][j]=(double *)malloc(dim3*sizeof(double ));
    }
    return array;
}// end double3d

void summ(double ***A,double ***B, double ****C)
{
    int i ,j ,k;
    for(i=0;i<10;i++)
    for(j=0;j<5;j++)
    for(k=0;k<5;k++)
    (*C)[i][j][k]=A[i][j][k]+B[i][j][k];
}

void main()
{
   int i,j,k,nx,ny;
   double ***M1, ***M2, ***M3, ***M4,***M5,***M6;
   nx=5;ny=5;
    M1=double3d(10,nx,ny);
    M2=double3d(10,nx,ny);
    M3=double3d(10,nx,ny);
    M4=double3d(5,nx,ny);
    M5=double3d(5,nx,ny);
    M6=(double ***)malloc(10*sizeof(double **));

    for(i=0;i<10;i++)
    {
        for(j=0;j<nx;j++)
           for(k=0;k<ny;k++)
           {
           M1[i][j][k]=i;
           M2[i][j][k]=1;
           }
    }

// Note random values are in M4 and M5 as they are not initalised
    memcpy(M6,     M4, 5 * sizeof(double **));
    memcpy(M6+5,   M5, 5 * sizeof(double **));

    for(i=0;i<5;i++)
    {
       for(j=0;j<nx;j++)
          for(k=0;k<ny;k++)
          {
               M4[i][j][k]=200;
               M5[i][j][k]=700;
          }
    }

printf(" printing M6 Memcpy before addtion\n");
    for(j=0;j<nx;j++)
    {
       for(k=0;k<ny;k++)
           printf("%f ",M6[4][j][k]);
       printf("\n");
       for(k=0;k<ny;k++)
           printf("%f ",M6[9][j][k]);
       printf("\n");
    }
    // calling for non memcpy array
    summ(M1,M2,&M3); printf(" Non memcpy output last value : %f \n",M3[9][nx-1][ny-1]);
    // calling for memcpy
    summ(M1,M2,&M6); printf(" memcpy output last value : %f \n",M6[9][nx-1][ny-1]);
printf(" printing M6 Memcpy for two sets after addtion\n");
    for(j=0;j<nx;j++)
    {
       for(k=0;k<ny;k++)
           printf("%f ",M6[4][j][k]);
       printf("\n");
    }
    for(j=0;j<nx;j++)
    {
       for(k=0;k<ny;k++)
           printf("%f ",M6[9][j][k]);
       printf("\n");
    }

    free(M6);// cleared M6

printf(" printing M4 Memcpy after deleting M6\n");
    for(j=0;j<nx;j++)
    {
       for(k=0;k<ny;k++)
           printf("%.1f ,%.1f ,%.1f ,%.1f ,%.1f ",M4[0][j][k],M4[1][j][k],M4[2][j][k],M4[3][j][k],M4[4][j][k]);
       printf("\n");
    }

printf(" printing M5 Memcpy after deleting M6\n");
    for(j=0;j<nx;j++)
    {
       for(k=0;k<ny;k++)
           printf("%.1f ,%.1f ,%.1f ,%.1f ,%.1f ",M5[0][j][k],M5[1][j][k],M5[2][j][k],M5[3][j][k],M5[4][j][k]);
       printf("\n");
    }
 }

如何在C中将数组拆分为两个数组

4 个答案: