segmentation fault (core dumped) gcc ubuntu - c

Question

I was trying to make a simple function to make a group of number that user enters them, using pointer of pointer but i keep getting this error and its hard to tell where the problem is, if there any other option to use something else that tells me where the problem is instead of this weird error.
#include <stdio.h>
void BuildGroub(int** group,int* count){
int i=0;
int j;
printf("Enter the size of the group \n");
scanf("%d", &*count);
while(*count != 0){
printf("Enter the %d number of the group:\n", i);
j=0;
scanf("%d", &**(group+i));
while(**(group+i)!=**(group+j)){
j++;
}
if(j==i){
i++;
count--;
} else{
printf("you have already entered this number please try again: \n");
}
}
}
int main(){
int count;
int group[100];
int *groupptr = &group;
BuildGroub(&groupptr,&count);
for(int i=0;i<count;i++){
printf("%d, ", group[i]);
}
return 0;
}

Answer 1

With this question, you do not need to use double pointer. If you want to learn how to use the double pointer, you can google then there are a ton of examples for you, for example, Double Pointer (Pointer to Pointer) in C.
In BuildGroub you decrease the count pointer
if(j==i){
i++;
count--;
}
, but in the condition of while loop, you compare the value that count pointer points to. it seems strange.
while(*count != 0)
Even if you change count-- to (*count)--, it will decrease the number of elements that you enter to 0 when you get out of the while loop, then in main function:
for(int i=0;i<count;i++){} // count is equal to 0 after calling BuildGroub function if you use (*count--) in while loop.
You should use a temp value for while loop function, for example:
int size = *count;
while(size != 0){
...
if (i == j) {
i++;
size--;
}
}
You should use, for example, group[i] instead of *(group+i). It will be easier to read your code.
The code complete:
#include <stdio.h>
void BuildGroub(int* group,int* count){
int i=0;
int j;
printf("Enter the size of the group \n");
scanf("%d", count);
int size = *count;
while(size != 0){
printf("Enter the %d_th number of the group:\n", i);
j=0;
scanf("%d", &group[i]);
while(group[i] != group[j]) {
j++;
}
if(j==i){
i++;
size--;
} else{
printf("you have already entered this number please try again: \n");
}
}
}
int main(){
int count;
int group[100];
int *groupptr = group;
BuildGroub(groupptr,&count);
for(int i=0;i<count;i++){
printf("%d, ", group[i]);
}
return 0;
}
The test:
./test
Enter the size of the group
5
Enter the 0_th number of the group:
1
Enter the 1_th number of the group:
2
Enter the 2_th number of the group:
2
you have already entered this number please try again:
Enter the 2_th number of the group:
3
Enter the 3_th number of the group:
3
you have already entered this number please try again:
Enter the 3_th number of the group:
4
Enter the 4_th number of the group:
5
1, 2, 3, 4, 5,

Answer 2

If you want to use a double pointer, you need to change your function like this:
void BuildGroub(int** group, int* count) {
int i = 0;
int j;
printf("Enter the size of the group \n");
scanf("%d", &*count); //I think this is redundant but works.
while (*count != 0) {
printf("Enter the %d number of the group:\n", i);
j = 0;
scanf("%d", (*group + i)); //The content of group + i
while ( *( *group + i) != *(*group + j)) { //the content of the content
j++;
}
if (j == i) {
i++;
(*count)--; //The content decrement
} else {
printf("you have already entered this number please try again: \n");
}
}
}
But you have a big problem in main and it is because you are using the parameter count to decrement until zero inside the function. So when the function finish, count value is zero and you don't print anything... You need to change this, using a internal variable to make the count, and finaly, setting the parameter to be using in main.