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Home/ Questions/Q 681197
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T01:27:26+00:00 2026-05-14T01:27:26+00:00

Consider the following code: int main() { int i; volatile int* p = &i;

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Consider the following code:

int main()                                                                      
{                                                                             
    int i;                                                                      
    volatile int* p = &i;                                                       
    int *v = p;                                                                 
    return 0;                                                                   
}

This gives an error in g++:

$ g++ -o volatile volatile.cpp 
volatile.cpp: In function ‘int main()’:
volatile.cpp:6: error: invalid conversion from ‘volatile int*’ to ‘int*’

My intention was that I want to make p volatile. However, once I’ve read the value of p, I don’t care if accessing v is volatile. Why is it required that v be declared volatile?

This is hypothetical code of course. In a real situation you could imagine that p points to a memory location, but is modified externally and I want v to point to the location that p pointed to at the time of v = p, even if later p is externally modified. Therefore p is volatile, but v is not.

By the way I am interested in the behaviour both when this is considered C and C++, but in C this only generates a warning, not an error.

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  1. Editorial Team
    Editorial Team
    2026-05-14T01:27:26+00:00Added an answer on May 14, 2026 at 1:27 am

    If you mean that the pointer should be volatile, rather than the object it points to, then declare it as

    int* volatile p;
    
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