Why there is an infinite for loop? - c

Question

The programm falls in the infinite loop. But arr->count printf prints a normal value (4, for example). count has a type unsigned int and arr is a pointer to int. What's the problem here? loop prints arr values at first and then continues to print trash values
In arrat_get it prints array just fine
struct _arr {
size_t count;
int* arr;
} ;
typedef struct _arr array_t;
array_t* array_get(FILE* file){
int* arr = NULL;
size_t count = 0;
array_t* arr_t;
array_t temp;
int i = 0;
if (!file) {
fprintf(stderr, "there is no such file\n");
return;
}
if (fscanf(file, "%u", &count) == EOF) {
fprintf(stderr, "can't read count from file\n");
return;
}
temp = array_create(arr, count);
arr_t = &temp;
printf("%i\n", arr_t->count);
for (i = 0; i < arr_t->count; i++){
if (fscanf(file, "%d", &arr_t->arr[i]) == EOF) {
fprintf(stderr, "can't read arr from file\n");
return;
}
}
for (i = 0; i<arr_t->count; i++)
printf("%d ", arr_t->arr[i]);
printf("\n");
return arr_t;
}
int main(){
array_t* arr_t;
int i = 0;
printf("enter count and arr:\n");
arr_t = array_get(stdin);
printf("count in main=%u\n", arr_t->count);
for (i = 0; i<arr_t->count; i++)
printf("%d ", arr_t->arr[i]);
getch();
return 0;
}

Answer 1

This is not the solution of your problem, but may help you to find it:
#include <stdio.h>
#include <stdlib.h>
struct _arr {
int count;
int* arr;
} ;
int main(){
int i = 0;
struct _arr myArray[10];
for (i = 0; i < 10; i++) {
myArray[i].count = 9;
myArray[i].arr = &myArray[i].count;
}
for (i = 0; i < (myArray->count); i++)
printf("%d %p\n", myArray[i].count, myArray[i].arr);
return 0;
}
OUTPUT:
Compiling the source code....
$gcc main.c -o demo -lm -pthread -lgmp -lreadline 2>&1
Executing the program....
$demo
10 0x7fffeb764c00
10 0x7fffeb764c10
10 0x7fffeb764c20
10 0x7fffeb764c30
10 0x7fffeb764c40
10 0x7fffeb764c50
10 0x7fffeb764c60
10 0x7fffeb764c70
10 0x7fffeb764c80