c - 在C中复制指针（矩阵）值的指针

Question

问题如下：我创建了一个动态矩阵，使用指针指针matrix1

我想将此矩阵的副本创建到另一个中，matrix2

我想这样做，这样我就可以在不搞砸的matrix2情况下搞砸matrix1 所以我尝试执行以下操作：

int main()
{
    int **matrix1, **matrix2, size1 = 10, size2 = 2;
    matrix1 = create_matrix(size1, size2);

    //I want to copy the value of matrix1 into matrixq2 and NOT the index
    **matrix2 = **matrix1
}

但是程序中断并显示以下内容：

我知道，从外观上看，使用该功能create_matrix两次会更容易， formatrix1和另一个 for matrix2。但是在我的原始程序中，这将是太多的工作，因为我做了很多事情来完成矩阵。哦，顺便说一句，我想避免使用 C++，有没有办法在不使用它的情况下做到这一点？对我来说会更好。

代码“create_matrix”如下：

//The program will read a file with the name of endereco, and create a matrix contPx3 out of it
int ** cria_matrix(char endereco[], int contP)
{
    FILE *fPointer;
    int i, contE, auxIndex, auxNum, **processos, cont_line = 0;
    char line[100];
    bool flag = true, flag2;

    fPointer = fopen(endereco, "r");

//Here the creation of the matrix
    processos = (int**)malloc(sizeof(int*) * contP);
    for (i = 0; i < contP; i++)
        processos[i] = malloc(sizeof(int) * 3);



//For now and on, is the rules of how the data will be placed on the matrix
    contP = 0;
    while (!feof(fPointer) && flag)
    {

        memset(&line[0], 'Ì', sizeof(line));
        fgets(line, 100 , fPointer);
//Bassicaly is that in each line there will be 3 numbers only, diveded but as many spaces you want. The numbeer will be placed on the matrix on the determined line they are.
        auxIndex = 0;
        flag2 = false;
        if(line[0] != '#')
            for (i = 0; i < 100; i++)
            {
                if (line[i] != ' ' && line[i] != '\n' && line[i] != '\0' && line[i] != 'Ì')//&&  line[i] != 'à'
                {
                    auxNum = line[i] - '0';
                    processos[contP][auxIndex] = auxNum;
                    auxIndex++;
                    flag2 = true;

                }
            }

        if (flag2)
            contP++;



        cont_line++;
        if (auxIndex != 3 && auxIndex != 0)
        {
            flag = false;
            printf("ERRO na linha: %d No processo: %d\nProvavelmente mais ou menos que 3 numeros separado por espacos\n", cont_line, contP);
        }

    }
    fclose(fPointer);
    if (!flag)
        system("PAUSE");
    return processos;
}

score 1 · Accepted Answer

这个怎么样——

matrix2 = (int**)malloc(sizeof(int*)*size1);
for(int idx = 0; idx < size1; ++idx) {
    matrix2[idx] = (int*)malloc(sizeof(int)*size2);
    for(int idx2 = 0; idx2 < size2; ++idx2) {
        matrix2[idx][idx2] = matrix1[idx][idx2];
    }
}

score 1 · Accepted Answer

matrix1指向行指针数组，*matrix1是指向保存第一行实际数据的数组的指针，并且**matrix1是第一行的第一个元素的值。matrix1它的每个元素都是动态分配的数组。

matrix2是您显示的代码中的未初始化（垃圾）指针。它既没有分配行指针也没有分配数据缓冲区。

为了达到你想要的结果，你需要先分配的元素matrix2，然后只复制 over 的数据部分matrix1。

int **copy_matrix(int **mat, int size1, int size1)
{
    int row;

    int **res = malloc(size1 * sizeof(int *));
    for(row = 0; row < size1; row++) {
        res[row] = malloc(size2 * sizeof(int));
        memcpy(res[row], mat[row], size2 * sizeof(int));
    }
    return res;
}

...

matrix2 = copy_matrix(matrix1, size1, size2);

另一种方法是为副本分配一个缓冲区。虽然这可能是一般存储矩阵的更好方法，但它可能对您没有那么有用，因为您将无法以matrix2与您相同的方式释放内存matrix1：

int **copy_matrix(int **mat, int size1, int size2)
{
    int row;
    int **res = malloc(size1 * sizeof(int *));
    res[0] = malloc(size1 * size2 * sizeof(int));

    for(row = 0; row < size1; row++) {
        res[row] = res[0] + row * size2;
        memcpy(res[row], mat[row], size2 * sizeof(int));
    }
    return res;
}

score -1 · Accepted Answer

您需要了解复制指针、浅拷贝和深拷贝之间的区别。考虑这个

struct employee
{
   char *name;
   float salary;
   int payrollid;
}

现在有三种复制员工的方式

struct employee * emp1;  // points to an employee, set up somehow
struct employee * emp2;  // empty pointer, null or whatever

emp2 = emp1;  // copy pointers. emp1 and emp2 now point to the same object.

指针拷贝

struct employee  employee1;  // employee, set up
struct employee  employee2;  // uninitalised emplyee

memcpy(&employee2, &employee1, sizeof(struct employee)); // shallow copy

浅拷贝

struct employee * emp1;  // points to an employee, set up somehow
struct employee * emp2;  // empty pointer, null or whatever

emp2 = copyemployee(emp1);

struct employee *copyemployee(struct employee *e)
{
   struct employee *answer = malloc(sizeof(struct employee));
   if(!answer)
     goto error_exit;
   answer->name = malloc(strlen(e->name) + 1);
   if(!answer->name)
      goto error_exit;
   strcpy(answer>name, e->name);
   answer->salary = e->salary;
   answer->payroolid = e->payrollid;
   return answer;
error_exit:
   /* out of memory handle somehow, usually by returning null
}

深拷贝

如您所见，即使对于只有一个可变长度字段的简单结构，进行深拷贝也是一项相当多的工作。都有它们的用途，尽管浅拷贝可能是三者中最没用且最容易出错的。

您可能只需要分配一个指针。

c - 在C中复制指针（矩阵）值的指针

3 回答 3

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