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Windows XP Professional x86 NT64 SP3 Cumulative Update for Intel® Client Technology Package (with [a new i[a]InfoAdvanced.v6.1.3.x86.NT64. For 32 bit and full version [Of Crack 1.20.7] All your files.Q:
Have a basic question on pointers
I have a really simple question and it’s really confusing me.
Imagine I have the following code.
int a = 5, b = 10;
int* p = &a;
int* q = &b;
int* p2 = &a;
*p2 = *p + 1;
When I use the statement p2 = *p + 1; the code executes in the following manner, right?
*p = 5 // before the statement
*p = 5 + 1 = 6 // p has become equal to a
*q = 10 // before the statement
*q = 10 + 1 = 11 // p2 has become equal to b
*p2 = 6 // so after the statement
Is that right?
A:
Yes, that’s right, except that it doesn’t “execute in the following manner” — it is literally exactly what you wrote:
int a = 5, b = 10;
int* p = &a;
int* q = &b;
int* p2 = &a;
*p2 = *p + 1;
It should be simple enough to see what happens, right? If you think about the order of operations, it becomes clear:
* p = 5 // p points to a
* q = 10 // p2 points to b
* p2 = (*p) + 1 // *p2 = 6
A:
Yes, that is correct.
What you did in the example (in terms of the assignment) is that you converted the memory address that p points to into a value that represents the address that q points to, which is the same address (since you did not assign q to anything else). The statement *p2
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