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o store the address of this function:void neyo(void);to the variable f of type pointer to a function that does not take any argument and does not return anything, you would do (check all correct answers if there is more than one):*f = neyo;*f = &neyo;f = &neyo;f = neyo;

Question

o store the address of this function:void neyo(void);to the variable f of type pointer to a function that does not take any argument and does not return anything, you would do (check all correct answers if there is more than one):*f = neyo;*f = &neyo;f = &neyo;f = neyo;

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Solution

The correct way to store the address of the function neyo to the variable f of type pointer to a function that does not take any argument and does not return anything is:

f = neyo;

This is because in C, the name of a function is a pointer to the function itself. So, you don't need to use the address-of operator &.

The other options are incorrect:

  • *f = neyo; is incorrect because *f is trying to dereference the function pointer, which is not what you want to do when simply assigning the function address to the pointer.
  • *f = &neyo; is incorrect for the same reason, and also because you don't need to use & with function names.
  • f = &neyo; is incorrect because, as mentioned, you don't need to use & with function names.

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