Array bounds infinite loop problem

 1 #include <stdio.h>
 2 
 3 int main()
 4 {
 5     int i = 0;
 6     int arr[3] = {0};
 7     for(; i<=3; i++){
 8         arr[i] = 0;
 9         printf("hello world\n");
10     }
11     return 0;
12 }

The code will run into an infinite loop of output hello world.

 

The reason is that array bounds, a [3] of the address specifies the address of the variable i.

It makes me wonder why a [3] addresses point to the address of the variable i?

 

 After interpretation of Gangster

 

 

 

 Just understand.

Because it will be 8-byte aligned address immediately behind the array will be i, when i = 3, i.e., the offset address of the array produce a [3] _address = base_address + 3 * type_size = i_address. Then i = 0, I sank into an endless loop.

 

 

For the 8-byte aligned, the same

 1 #include <stdio.h>
 2 
 3 int main()
 4 {
 5     int j=3;
 6     int i=0;
 7     int arr[3] = {0};
 8     for(; i<=3; i++){
 9         arr[i] = 0;
10         printf("%d\n",i);
11         printf("hello world\n");
12     }
13     printf("%d%d\n",j,i);
14     return 0;
15 }

#include <stdio.h>

int main ()
{
    int i = 0;
    int j=3;
    int arr[2] = {0};
    for(; i<=3; i++){
        arr[i] = 0;
        printf("%d%d\n",j,i);
        printf("hello world\n");
    }
    
    return 0;
}