以抽象数据类型打印矩阵的函数

Question

矩阵中。 c 我有

struct matrix{
    int row;
    int col;
    int *a;
};

Matrix* allocateMemory(int m, int n) {
    Matrix* mat = (Matrix*) malloc(sizeof(Matrix));

    if (mat == NULL) {
    return NULL;
}
mat->row = m;
mat->col = n;
mat->a = (int*)calloc(m*n, sizeof(int));
return m;
}

虽然在matrix.h我有

#ifndef MATRIX_H_INCLUDED
#define MATRIX_H_INCLUDED

typedef struct matrix Matrix;

Matrix* allocateMemory(int m, int n);  
//...
int printMatrix(Matrix* mat); 

int transpose(Matrix* mat); 

#endif // MATRIX_H_INCLUDED

我正在使用矩阵的 ADT，但我坚持使用打印功能。我想创建一个函数

int printMatrix(Matrix* mat)

所以在 main.c 我会有类似

int i, j;
for (i = 0; i < row; i++){
    for(j = 0; j < col; j++){           
        printf("%d    ", printMatrix(mat));
    }
    printf("\n");
}

这意味着我想在 main 中而不是在函数中使用 printf，但我可以做到这一点

void printMatrix(Matrix* mat){
int i, j, k;
    for (i = 0; i < mat->row; i++){
        for(j = 0; j < mat->col; j++){
            k = i*mat->col + j;
            printf("%d   ", mat->a[k]);
        }
        printf("\n");
    }   
}

它确实打印了矩阵，但它似乎不正确。转置矩阵函数也是一样的，它确实正确地打印了转置矩阵，但我想要一个

int transpose(Matrix* mat)

所以我会在 main 中使用函数 printMatrix 来打印转置，但我只能这样做

void transpose(Matrix* mat){
int i, j, k;
    for (i = 0; i < mat->col; i++){
        for (j = 0; j < mat->row; j++){
            k = j*mat->row + i;
            printf("%f     ", mat->a[k]);
        }
        printf("\n");
    }    
}

How can I create the int function to print the matrix?

I am still studying ADT, but what would be my lack of understanding so I couldn't do the function?

Answer 1

这是我的热门评论。

更多风格提示...

不要不要投malloc return [等。等]

更惯用的是（例如）：

Matrix *mat = malloc(sizeof(*mat));

我知道在学校里，他们教授使用（例如）i, j, k，但尝试使用更具描述性的名称（例如）row, col, off.

并且，使参数也具有描述性：

Matrix *allocateMemory(int row,int col)

这是重构后的版本[进行了一些样式清理]：

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

typedef struct matrix {
    int row;
    int col;
    int *a;
} Matrix;

Matrix *
allocateMemory(int row, int col)
{
    Matrix *mat = malloc(sizeof(*mat));

    if (mat == NULL) {
        perror("malloc");
        exit(1);
    }

    mat->row = row;
    mat->col = col;

    mat->a = calloc(row * col, sizeof(*mat->a));
    if (mat->a == NULL) {
        perror("calloc");
        exit(1);
    }

    return mat;
}

void
printMatrix(Matrix *mat)
{
    int row, col, off;

    for (row = 0; row < mat->row; row++) {
        for (col = 0; col < mat->col; col++) {
            off = (row * mat->col) + col;
            printf("%d   ", mat->a[off]);
        }
        printf("\n");
    }
}

void
matrix_fill(Matrix * mat)
{
    int row, col, off;
    int val = 1;

    for (row = 0; row < mat->row; row++) {
        for (col = 0; col < mat->col; col++) {
            off = (row * mat->col) + col;
            mat->a[off] = val++;
        }
    }
}

void
transpose_copy(Matrix *matout,Matrix *matin)
{
    int row, col;
    int inoff, outoff;

    for (row = 0; row < matin->row; row++) {
        for (col = 0; col < matin->col; col++) {
            inoff = (row * matin->col) + col;
            outoff = (col * matout->col) + row;
            matout->a[outoff] = matin->a[inoff];
        }
        printf("\n");
    }
}

Matrix *
transpose_new(Matrix *matin)
{
    Matrix *matout = allocateMemory(matin->col,matin->row);

    transpose_copy(matout,matin);

    return matout;
}

int
main(void)
{

    Matrix *matin = allocateMemory(2,3);
    matrix_fill(matin);

    printf("Original:\n");
    printMatrix(matin);

    Matrix *matout = transpose_new(matin);

    printf("Transposed:\n");
    printMatrix(matout);

    return 0;
}

程序输出如下：

Original:
1   2   3
4   5   6

Transposed:
1   4
2   5
3   6

以抽象数据类型打印矩阵的函数

Function to print Matrix in abstract data type

c

abstract-data-type