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Home/ Questions/Q 7563053
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T13:34:02+00:00 2026-05-30T13:34:02+00:00

This is the code: char *(*strcpy_ptr)(char *dst, const char *src); Pointer to strcpy-like function

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This is the code:

char *(*strcpy_ptr)(char *dst, const char *src); Pointer to strcpy-like function

And the tutorial says:

Note the parentheses around *strcpy_ptr in the above declaration.
These separate the asterisk indicating return type (char *) from the
asterisk indicating the pointer level of the variable (*strcpy_ptr —
one level, pointer to function).

I’m lost on this – where is the “function pointer” and what is the “pointer level” ?

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  1. Editorial Team
    Editorial Team
    2026-05-30T13:34:03+00:00Added an answer on May 30, 2026 at 1:34 pm

    You are declaring a variable strcpy_ptr. You want this variable to be a pointer to a function returning a char*. If you did it without the parentheses this way:

    char **strcpy_ptr(char *dst, const char *src);
    

    It would be the prototype of a function that returns a char** – not what you want. The parentheses are to group one star with the variable, and seperate the star from the return type.

    Remember that pointers are declared like this:

    T *var;
    

    Where T is some type. The more stars you add, the more levels of indirection you add before you finally get to the actual T. So char **c would be a pointer to a pointer to a char. It’s the same thing for function pointers: T is char*, and *var must be seperated by parentheses, because C is ignorant of white space. C just added a little extra syntax to specify what kind of and how many arguments the function takes that is pointed to by the pointer. This is just part of the way C works.

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