Question

我想在点人的帮助下扫描一个2D阵列,并撰写了这一法典,请告诉我,编辑者为什么会犯错误? I know How to use Double pointers to do same, i is conducted with this.

#include<stdio.h>
#include<stdlib.h>
int main(void) {
    int i,j,n,a,b;
    int (*(*p)[])[];
    printf("
	Enter the size of the matrix in the form aXb	
");
    scanf("%dX%d",&a,&b);
    p=(int (*(*p)[b])[a])malloc(b*sizeof(int (*p)[a]));
    for(i=0;i<b;i++) {
            p[i]=(int (*p)[a])malloc(a*sizeof(int));
            printf("	Enter Column %d	
");
            for(j=0;j<a;j++)
                    scanf("%d",&p[i][j]);
    }
    return 0;
}

Answer 1

你们正在使用点数阵列,因此,你应直接将其索引,因为<代码>p[i]将*(p+i),即,在页注之后的阵列,而不是第1页的内容。

在C中,ave*将转换成任何点类型,因此,你无需得出小区的结果。如果你确实把 cast子放在一边,它就能够掩盖错误,例如,如果你试图指派给非点人(例如<代码>p[i])。

在花名册上,sizeof(int (*p) [a] 要么使用某种类型,要么使用某种表述,要么使用一种声明。 <代码>p是一系列星体的指点器,因此<代码>*p各项要素的类型为。

因此,这汇编了无错误的或关于电离的警告:

#include<stdio.h>
#include<stdlib.h>
int main ( void )
{
    int i, j, n, a, b;

    int ( * ( * p ) [] ) [];

    printf ( "
	Enter the size of the matrix in the form aXb	
" );

    scanf ( "%dX%d", &a, &b );

    p = malloc ( b * sizeof ( int ( * ) [] ) );

    for ( i = 0;i < b;i++ ) {
        ( *p ) [i] = malloc ( a * sizeof ( int ) );
        printf ( "	Enter Column %d	
", i );
        for ( j = 0;j < a;j++ )
            scanf ( "%d", & ( * ( *p ) [i] ) [j] );
    }


    return 0;
}

然而,由于在使用一个阵列的点子反对使用其第一个要素的点子方面没有任何好处,但是这确实意味着,在采用该要素之前,你必须加以参照,因此使用点子更容易使用。

Answer 2

Do you know what int (*(*p)[])[] is?
Try cdecl.org ... http://cdecl.ridiculousfish.com/?q=int+%28%2A%28%2Ap%29%5B%5D%29%5B%5D

a. 使用1维阵列,并预设2维阵

declare a 1-dimensional object (pointer, array, whatever)
malloc a rectangular size
compute linear addressing value based on row, column, and column size
use it
free the array

......

/* Oh ... and use spaces in your code */
/* They are extremely cheap now a days */
#include <assert.h>
/* instead of asserting malloc and scanf, use proper error checking */
#include <stdio.h>
#include <stdlib.h>

int main(void) {
    int i, j, n, rows, cols;
    int *p;                                            /* 1. */

    printf("Enter the size of the matrix in the form aXb
");
    n = scanf("%dX%d", &rows, &cols);
    assert((n == 2) && ("scanf failed"));
    p = malloc(rows * cols * sizeof *p);               /* 2. */
    assert((p != NULL) && "malloc failed");
    for (i = 0; i < rows; i++) {
            int rowindex = i * cols;                   /* 3. */
            for (j = 0; j < cols; j++) {
                    n = scanf("%d", &p[rowindex + j]); /* 3. and 4. */
                    assert((n == 1) && "scanf failed");
            }
    }
    free(p);                                           /* 5. */
    return 0;
}

Answer 3

You are unnecessarily complicating the problem of accessing array elements using pointers. Try to use a simple pointer-to-pointer p.

int **p;
...
p=malloc(a*sizeof(int *));   //create one pointer for each row of matrix
...
for(i=0;i<a;i++)
{
...
p[i]=malloc(b*sizeof(int));  //create b integers in each row of matrix
...
}

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