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use MALLOC function on array

zhu_zheng
zhu_zheng asked
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Last Modified: 2010-04-15
Greetings,

I am trying to use MALLOC function to set the memory location for array which is defined by pointer to pointer. The code doesn't has error, but it couldn't run. The part of code is attached as following:

int i,j;
int **matrix;

main()
{
  matrix=(int**)malloc(5*5*sizeof(int));
   
  /*initialize the array.*/
  for (i=0; i<=4; i++)
   for(j=0;j<=4;j++)
    *(*(matrix+i)+j)=0;
}

Is there anything wrong with the usage of MALLOC? The code stop at
*(*(matrix+i)+j)=0

-Zheng
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ozo
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Commented:
 matrix=(int**)malloc(5*sizeof(int*));

  /*initialize the array.*/
  for( i=0; i<=4; i++ ){
      matrix[i] = (int *)malloc(5*sizeof(int));
      for(j=0;j<=4;j++){ *(*(matrix+i)+j)=0; }
  }
In your program, try

  (int *)matrix= malloc(5*5*sizeof(int));
Commented:
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ozo's solution works.

ufolks123's solution probably works because it is very similar to ozo's.

Zheng, your way does not work because you have not informed the compiler that matrix has the form of a [5][5] array, so it does not know what units i is specified in when you say "matrix + i".


For interest, here is a solution whose main advantage is that it uses a single malloc:

#include "malloc.h"
#define NUM_OF_ROWS      5

typedef int    matrixRow[5];

main()
{
  int i,j;
  matrixRow  *matrix;

  matrix = malloc(NUM_OF_ROWS * sizeof(matrixRow));

  /*initialize the array.*/
  for (i=0; i<=NUM_OF_ROWS; i++)
   for(j=0;j<=4;j++)
    *(*(matrix+i)+j)= 0;
}

This could be changed to use ufolk123's [][] matrix addressing if required.
In the previous post, change
i<=NUM_OF_ROWS   to i<NUM_OF_ROWS

Author

Commented:
Thank you very much for you inputs.
-Zheng
Hi zhu_zheng

Welcome to Experts Exchange.

I am interested to know why you chose ufolks answer rather than that of another contributor.
Was it the first one that worked?
Was it the best one?
Was it highly original or elegant?

You should not necessarily award the points to the first person who posts an "answer".  You choose the one that YOU want.

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