Put array as argument and read its value - c

Question

I am scanning for values using function .
int **array(int * counter) {
int **vrat;
int max = 5;
int index = 0;
int i;
vrat = malloc(max*sizeof(int*));
for ( i = 0; i < max ; i++) {
vrat[i] = malloc(2 * sizeof(int));
}
int x;
int y;
char c;
while (scanf("%d%c%d", &x, &c, &y) != EOF) {
vrat[index][0] = x;
vrat[index][1] = y;
index++;
}
*counter = index;
return vrat;
}
and calling it in main to return the array, which works .
int main()
{
int counter=0;
int **mej;
int gg;
mej = array(&counter);
int i;
gg = pocet(mej, &counter);
return 0;
}
what is bothering my mnd is "pocet" function , i am passing an array in it and want to print its value . but it always print undefined numbers
Function looks like this
int pocet(int array[][2],int *counter) {
int poc = 0;
int i;
for ( i =0; i < *counter ;i++) {
printf("cislo = %d", array[i][0]);
}
return poc;
}
as you can see , it has static 2nd dimension , how can i make this work?

Answer 1

You are allocating space for 5 pointers, you should multiply by the size of a pointer and not the size of an int, like this
vrat = malloc(max * sizeof(int *));
/* ^ pointer */
Although you can completely avoid the mistake by multiplying by the size of the pointer type like this
vrat = malloc(max * sizeof(*vrat));
and ALWAYS check that malloc() has not returned NULL before actually using the pointer.
Also, don't compare scanf() to EOF since it's very unlikely to get an EOF before an input error because it requires explicit input from the user, instead do this
while(scanf("%d%c%d", &x, &c, &y) == 3)
The return type of your function should match that of the returned object, in this case int **, change the function signature to
int **array(int *content)

Answer 2

vrat = malloc(max*sizeof(int*));
How do you know whether the memory successfully got allocated or not?
Always check the pointer returned by malloc & family functions for equivalence with NULL to avoid possible SegFault.
while (scanf("%d%c%d", &x, &c, &y) != EOF)
What if there's matching failure for c? Then scanf will return 1 and while will still go for iteration even if scanf could not store values in c and y.
So always compare scanf return value with no. of arguments instead of EOF
while (scanf("%d%c%d", &x, &c, &y) != EOF) {
vrat[index][0] = x;
vrat[index][1] = y;
index++;
}
Here what if index >= max?
Since you have allocated memory for only max integer pointers you must not try to access the memory beyond the allocated chunk.
Solution: Change while condition to
while ((scanf("%d%c%d", &x, &c, &y) == 3) && (index < max))
So the correct code snippet would be:
vrat = malloc(max*sizeof(int*));
if(!vrat)
{
printf("vrat: malloc failed!\n");
exit(1);
}
for ( i = 0; i < max ; i++) {
vrat[i] = malloc(2 * sizeof(int));
if(!vrat[i])
{
printf("vrat[%d]: malloc failed!\n", i);
exit(1);
}
}
int x;
int y;
char c;
while ((scanf("%d%c%d", &x, &c, &y) == 3) && (index < max)) {
vrat[index][0] = x;
vrat[index][1] = y;
index++;
}

Answer 3

An int** is not an int[][2]. The first is a pointer to pointers to int while the second (as an argument) is a pointer to int[2]. You cannot convert between the two the way you tried. You should have gotten a compiler warning/error about this.
The easiest way to fix this is to have pocet take an int** as the first argument. Fix the issues others have mentioned as well.

Answer 4

You have defined int **vrat;
When you do malloc, you need to use int ** or int * and not int