CODEWITHSUNDEEP
cpointers
Eslam Hamdy
Eslam Hamdy

Reputation: 7386

Pointer to Pointer to integer

i'm using a pointer to pointer to intger as a 2 Dims array, i wrote these code but i failed on getting the integer value.

#include<stdio.h>
#include<conio.h>

void main(void)
{
    int **pptr = 0, size = 0, size2 = 0, i, j;

    clrscr();
    printf("Enter Size of Main Store n");
    scanf("%d", &size);

    pptr = (int **) malloc(size * sizeof(int *));

    printf("Enter Size of Sub Store n");
    scanf("%d", &size2);

    for (i = 0; i < size; i++) {
        pptr[i] = (int *) malloc(size2 * sizeof(int));
    }

    printf("Enter Values n");

    for (i = 0; i < size; i++) {
        for (j = 0; j < size2; j++) {
            scanf("%dn", pptr[i][j]);
        }
    }

    clrscr();
    printf(" Valuesn");

    for (i = 0; i < size2; i++, pptr++) {
        printf("%dn", *pptr + i);
    }

    getch();
}

it prints rubbish!!

Upvotes: 0

Views: 161

Answers (2)

Daniel Fischer
Daniel Fischer

Reputation: 183883

scanf("%d", arg) expects a pointer to int, but

for (i = 0; i < size; i++) {
    for (j = 0; j < size2; j++) {
        scanf("%dn", pptr[i][j]);
    }
}

you pass it an int, an uninitialised int at that. The indeterminate value in that memory location is then interpreted as a pointer, and the scan tries to store the converted value who-knows-where. It is not unlikely that that will cause a segmentation fault.

You should pass &pptr[i][j] as the argument there.

for (i = 0; i < size2; i++, pptr++) {
    printf("%dn", *pptr + i);
}

prints the int* *pptr +i, which is &pptr[0][i], using the %d format that expects an int argument.

You should

printf("%d\n", pptr[0][i]);

if you want to print the values of the diagonal (as seems to be the case, since you also increment pptr in the loop), or, better

for(i = 0; i < size && i < size2; ++i) {
    printf("%d\n", pptr[i][i]);
}

or, if you want to print the entire grid,

for(i = 0; i < size; ++i) {
    for(j = 0; j < size2; ++j) {
        printf("%d ", pptr[i][j]);
    }
    printf("\n");
}

Upvotes: 2

Lee
Lee

Reputation: 13542

scanf takes a int *, not an int. So...

scanf("%dn", pptr[i][j]);

should be:

scanf("%dn", &(pptr[i][j]));

or it could be:

scanf("%dn", (pptr[i]+j));

also, printf needs a value (int), not a pointer. So...

printf("%dn", *pptr + i);

should probably be something like:

printf("%dn", *(*pptr+i));

or the much nicer looking equivalent:

printf("%dn", pptr[0][i]);

Upvotes: 0

Related Questions