Passing an array to a pointer - c

Question

I'm trying to learn C Language using pointers and arrays etc... but my code won't work. Whats is wrong with this code?
#include <stdio.h>
#include <conio.h>
#include <windows.h>
#include <stdlib.h>
#include <time.h>
int main(){
int *number = 0;
int i = 0;
int *random = 0;
int randomTry[100];
int *getRandomTry = 0;
int randomGenerator[100];
int *getRandomGenerator = 0;
srand (time(NULL));
*getRandomTry = randomTry[100];
*getRandomGenerator = randomGenerator[100];
do{
*random = rand() % 10;
getRandomGenerator = random;
printf("Choosing a number:\n\r");
for(i = 0; i < 30; i++){
printf("*");
Sleep(50);
}
printf("\n\r\n\rWell done, your time.");
printf("\n\r------------------------------");
printf("\n\rPick a number ( 0 to 10): ");
scanf("%d", number);
getRandomTry = number;
if(*number > 10){
printf("\n\rRemember, number only from (0 to 10)\n\r");
}
else{
if(number == random){
printf("\n\rYou choose right!");
printf("You entered theses numbers till the right answer");
for(i = 0; i < randomTry[i]; i++){
printf("%i", getRandomTry);
}
}
else{
printf("\n\rYou choose wrong, number: %d - random: %d\n\r", number, random);
Sleep(700);
system("cls");
}
}
}while(number != random);
getch();
return 0;
}
When i compile the program stop working, no warnings or comments show in console.
I'm using the notepad++ with MinGw to compile this code.

Answer 1

Assigning a value to a NULL pointer is undefined behaviour, such as (in your code).
int *getRandomTry = 0;
*getRandomTry = randomTry[100];
gives undefined behaviour on two counts. The second statement tries to retrieve the value of randomTry[100], which doesn't exist (since array indexing starts at zero, and randomTry only has 100 elements). Second, it tries to store that value into a location addressed by the NULL pointer.
With undefined behaviour, anything can happen. Program crashes are pretty common, but not the only possible result.
So you need to understand how array indexing works (it starts at zero, not one). And you need to ensure all pointers point at something valid before trying to set or retrieve values via the pointer.
For example;
int x = 5;
int *getRandomTry = &x;
*getRandomTry = 42; // will change x

Answer 2

The following is a correct declaration of a pointer to an integer, with initial value zero:
int *random = 0;
The following statement will try to place the result of the calculation to the address in memory where the pointer 'random' is pointing:
*random = rand() % 10;
Since 'random' was assigned the value of 0, it is pointing at memory location zero, and the second statement will result in 'undefined behavior', meaning that what happens next in your program is not easily predictable. In practice, it will most likely crash in one way or another, since you cannot write to memory location 0.
if the code was like this:
int random = 0;
...
random = rand() % 10;
then everything would work fine. It would also work if code was like this:
int random_variable;
int *random = 0;
...
random = &random_variable;
...
*random = rand() % 10;
In this last case, 'random_variable' would end up with the result of the calculation.
Hope this helps.