int *p = &a + 2;
In this line, you declare a pointer variable p of type int*. The &a takes the address of variable a and + 2 increments the address by two int sizes. So, p now points to the memory location that is two int sizes ahead of a.
2. *p = 22;
Here, you assign the value 22 to the memory location pointed by p. Since p was set to be two int sizes ahead of a, this statement actually assigns the value 22 to the memory location two int sizes ahead of a.
After analyzing the code, we can conclude that:
Initially, the value of a is uninitialized and can hold any arbitrary value.
The pointer p is set to point two int sizes ahead of a.
The statement *p = 22; assigns the value 22 to the memory location pointed by p. Since p points two int sizes ahead of a, the assignment actually modifies the value at that location, not the value of a itself.
Considering the order of the statements, if you execute the code as it is, without initializing a before these statements, you would encounter undefined behavior. This means the output of printf("%d", a); cannot be predicted reliably.
However, if you were to initialize a before the code snippet, the output of printf("%d", a); would be 100 because the value of a would not be affected by the assignment *p = 22;.
#Programming #Pointers #CodeAnalysis #TechInsights #AbdurRahimRatulAliKhan
