deferencing the int malloc pointer - c

Question

int main()
{
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int rslt, n;
float sum = 0;
scanf("%d", &n);
int *tri = (int*)malloc(n * 3 * sizeof(int));
if(tri == NULL){
return 1;
}
printf("%d\n", *(tri[0]));
when I am referencing the tri pointer then it is showing the error.
invalid type argument of unary '*'.
thanks for the answer.
The above part is clarified but I have another issue. but I have another issue
int main()
{
int rslt, n;
float sum = 0;
scanf("%d", &n);
int *tri = (int*)malloc(n * 3 * sizeof(int));
if(tri == NULL){
return 1;
}
for (int i = 0; i < n; i++)
{
for (int j = 0; j < 3; j++)
{
scanf("%d", &(tri[j]));
}
printf("\n");
}
for (int i = 0; i < n; i++)
{
for (int j = 0; j < 3; j++)
{
sum += tri[j] / 2;
printf("%d %d\n",sum,tri[j] / 2);
}
}
printf("%d\n",sum);
return 0;
}
when I am printing the tri[j] it is showing some garbage value.

Answer 1

The variable tri has the type int * due to this declaration
int *tri = (int*)malloc(n * 3 * sizeof(int));
Pay attention to that you allocated an uninitialized memory.
The expression tri[0] has the type int. And you are trying to apply the dereference operator to the expression of the type int
*(tri[0])
So the compiler issues an error.
Instead you could write
int *tri = (int*)calloc(1, n * 3 * sizeof(int));
and then write either
printf("%d\n", tri[0]);
or
printf("%d\n", *tri);
These for loops
for (int i = 0; i < n; i++)
{
for (int j = 0; j < 3; j++)
{
scanf("%d", &(tri[j]));
}
printf("\n");
}
for (int i = 0; i < n; i++)
{
for (int j = 0; j < 3; j++)
{
sum += tri[j] / 2;
printf("%d %d\n",sum, tri[j] / 2);
}
}
printf("%d\n",sum);
are incorrect. There are used an incorrect conversion specifier %d instead of %f with the variable sum and incorrect expressions for indices. You need to write
for (int i = 0; i < n; i++)
{
for (int j = 0; j < 3; j++)
{
scanf("%d", &(tri[3 * i + j]));
}
printf("\n");
}
for (int i = 0; i < n; i++)
{
for (int j = 0; j < 3; j++)
{
sum += tri[3 * i + j] / 2;
printf("%f %d\n",sum, tri[3 * i + j] / 2);
}
}
printf("%f\n",sum);

Answer 2

The array expression a[i] is defined as *(a + i) - given some address value a, offset i elements (not bytes!) from that address and dereference the result, so
*tri == *(tri + 0) == tri[0]
The expression tri[0] has type int, which is why the compiler is complaining when you write *(tri[0]); the operand of the unary * operator must have a pointer type.
So in your printf statement you would use either *tri or tri[0]. The type of either expression is int.
As others have said, malloc does not initialize the allocated memory, so the value of tri[0] may be anything. You could use calloc instead, which initializes the memory to all-bits-zero:
int *tri = calloc( n, sizeof *tri );
In C you do not need to cast the result of malloc, calloc, and realloc1, and most of us will advise you to not do so. Similarly, since the type of the expression *tri is int, the result of sizeof *tri is the same as sizeof (int). This way you don't have to worry about keeping types straight between the declaration, the cast, and the sizeof expression, so you're less likely to make a mistake.
This is not the case in C++, but if you're writing C++ you shouldn't be using the C *alloc functions anyway.