C: Array output

Question

Given an integer array, I'm trying to print the array with all the zeros moved to the left of the array. The order of the remaining numbers doesn't matter. I keep getting strange outputs like "{-1073741824,1049472,1,49,63,1055984,1}" for the array that is hardcoded in main.

int main(int argc, const char * argv[]) {
    int a[10] = {3, 0, 1, 4, 0, 0, 7, 20, 1, 5};
    int n = 10;
    int count = 0;
    for (int i = 0; i < n; ++i)
    {
        if (a[i] == 0)
        {
            ++count;
        }
    }

    //////////

    int *array = malloc(0);
    for (int j = 0; j < count; ++j)
    {
        array = realloc(array, (j + 1) * sizeof(int));
        array[j] = 0;
    }

    //////////

    printf("%s", "{");
    for (int k = 0; k < n-1; ++k)
    {
        if (array[k] != 0)
        {
            printf("%d%s", array[k], ",");
        }
    }
    printf("%d", array[n-1]);
    printf("%s", "}\n");

    //////////

    free(array);
    return 0;
}

So, you define a as an array. Then you define another array, named array. Then you use array. But where do you use or copy the values from a? — AndASM
– AndASM, Commented Feb 8, 2016 at 23:43
malloc(0) is undefined behavior... Though you've certainly used it in an interesting way. :) I doubt it's the source of your problem. Take note that you may actually have a valid pointer returned from that of a minimum allocation size (8 or 16 bytes). It's also a super good idea to check that malloc and realloc return valid pointers any time they are called. — David Hoelzer
– David Hoelzer, Commented Feb 8, 2016 at 23:45
Thanks AndASM, I realized that I'm printing from the wrong array. I'll see if I can fix it! — TyCharm
– TyCharm, Commented Feb 8, 2016 at 23:49
I suggest you learn some basic debugging skills either with a source level debugger or with well-placed printf() statements. — Code-Apprentice
– Code-Apprentice, Commented Feb 8, 2016 at 23:59

Martín Muñoz del Río · Accepted Answer · 2016-02-08 23:57:31Z

2

You can replace:

int *array = malloc(0);
for (int j = 0; j < count; ++j)
{
    array = realloc(array, (j + 1) * sizeof(int));
    array[j] = 0;
}

With something like:

int array[10]; //malloc(0);
int j = 0;
for (j = 0; j < count; ++j)
{
    array[j] = 0;
}

for (j = 0; j < n; ++j)
{
    if(a[j]!=0)
        array[count++] = a[j];
}

If you use this code you don't need malloc, realloc neither free.

answered Feb 8, 2016 at 23:57

Martín Muñoz del Río

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Comments

user2612740 · Accepted Answer · 2016-02-09 23:03:33Z

0

The code looks overly complicated - you don't need to pass a string in using %s you can just print it directly. So

printf("%s", "{");

can be:

print("{");

After your first step, you could just print the number of zeros you've found, followed by stepping through and printing all the non-zero integers. Something a bit like this

int a[10] = {3, 0, 1, 4, 0, 0, 7, 20, 1, 5};
int n = 10;
int count = 0;
for (int i = 0; i < n; ++i)
{
    if (a[i] == 0)
    {
        ++count;
    }
}
printf("{");
for(int i=0; i < count; i++)
{
     printf("0,");
}
for(int i=0; i < n; i++)
{
     if (a[i]!=0) {
         printf("%d", a[i]);
         if (i < n-1) {
             printf(",");
         }
     }
}
printf("}\n");

edited Feb 9, 2016 at 23:03

answered Feb 9, 2016 at 0:13

user2612740

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