i am facing problems while running a code on GNU GCC compiler while it's working fine on VC++ compiler

Question

printf("\n enter your choice\n 1 to ask user for the size of game board\n 2 tp get input from a file\n");
scanf("%d",&choice);
if(choice==1)
{
    printf("\n enter number of rows and columns");
    scanf("%d%d",&row,&col);
    mat=(char**)malloc(row*sizeof(char));
    label=(char**)malloc(row*sizeof(char));
    for(i=0;i<row;i++)
    {
        mat[i]=(char*)malloc(col*sizeof(char));
        label[i]=(char*)malloc(col*sizeof(char));
    }

    for(i=0;i<row;i++)
    {

        for(j=0;j<col;j++)
        {
            temp=rand()%5;

            mat[i][j]=color_codes[temp];
            label[i][j]=' ';
        }

    }
}

this C Language statements is running fine on my visual studio IDE, but when i try to run this same statements on CODE BLOCKS IDE this thing suddenly crashes .Any kind of help will be highly appreciated

Yes, and you need to say which part it crashes at. Are we just supposed to magically know? — SevenBits, Jun 09 '13 at 14:49
And you need to further tell us the declarations of **all variables**. — Jens, Jun 09 '13 at 14:55
i think Seven bits you should see a answer below which might make you realise that knowledge comes from experiences — user2468471, Jun 09 '13 at 15:59

score 2 · Accepted Answer · edited May 23 '17 at 11:58

2

There's an easy way to avoid this bug, where you multiply with the wrong value, when doing a malloc:

→ Always multiply with the sizeof the type of the array element, sizeof(*var) ←.

 char **mat;

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

This way you'll never ever get the multiplications wrong again. Another advantage: Should you decide to change the type of mat to say, double, all you need to change is one place: its declaration.

And you shouldn't cast the malloc return value. It's been hashed to death on SO why.

edited May 23 '17 at 11:58

Community

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answered Jun 09 '13 at 15:00

Jens

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+1 for the array variable in sizeof and for throwing unnecessary casts away. – A4L Jun 09 '13 at 15:03

score 1 · Answer 2 · answered Jun 09 '13 at 14:49

1

You are not allocating enough memory:

mat=(char**)malloc(row*sizeof(char));
label=(char**)malloc(row*sizeof(char));

Each item in the arrays is a pointer to char (char*), so you should multiply by sizeof(char*):

mat=(char**)malloc(row*sizeof(char*));
label=(char**)malloc(row*sizeof(char*));

answered Jun 09 '13 at 14:49

nullptr

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And you **shouldn't cast** the malloc return value. It's been hashed to death on SO why. – Jens Jun 09 '13 at 14:57
thnks buddy really helpful – user2468471 Jun 09 '13 at 15:06

A4L · Answer 3 · 2013-06-09T15:10:21.370

1

the two lines

mat=(char**)malloc(row*sizeof(char));
label=(char**)malloc(row*sizeof(char));

look not sane to me, you are trying to allocate an array of pointers to char but you specify only the size for a simple char. a pointer usually needs more that one byte.

so what you probably need is

mat=(char**)malloc(row*sizeof(char*));
label=(char**)malloc(row*sizeof(char*));

Note that it is not necessary to cats the result of malloc since in C a void* can be implicitly casted to any type.

So

mat=malloc(row*sizeof(char*));
label=malloc(row*sizeof(char*));

will work fine too.

edited Jun 09 '13 at 15:10

answered Jun 09 '13 at 14:51

A4L

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And you **shouldn't cast** the malloc return value. It's been hashed to death on SO why. – Jens Jun 09 '13 at 15:01

i am facing problems while running a code on GNU GCC compiler while it's working fine on VC++ compiler

3 Answers3