What is the value of n after the following code is executed?int n = 98;int *p = &n;*p++;
Question
What is the value of n after the following code is executed?
int n = 98;
int *p = &n;
*p++;
Solution
The code you provided is incorrect and will not compile. The line "*p++;" is trying to increment the value of the pointer 'p', not the value 'n' that 'p' points to. If you want to increment the value of 'n' using the pointer 'p', you should use "(*p)++;" instead.
However, if the code was "*p++;", it would not change the value of 'n'. It would instead increment the memory address that 'p' is pointing to, which is not what you want.
So, if you correct the code to:
int n = 98;
int *p = &n;
(*p)++;
The value of 'n' after the code is executed would be 99.
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