c - How to understand dereferences and generic pointer in order to see the output? -

why output: aa b ee f ii j? did line: void (*pfunc)(void*); create function pfunc similar function print? know void pointer generic pointer still, how did output come way is? arguments @ command line: aaa eee iii

void print (void *a) {    char**p=(char**)a;     printf("%s",(*p+1));    putchar('');    putchar(**p+1);    putchar('');  } int main(int argc, char *argv[]) {    int i;    void(*pfunc)(void*);    pfunc=print;    for(i=1; i< argc;i++)       pfunc(argv+i);    return 0; } 

did row: void (*pfunc)(void*); creates function pfunc similar function print?

yes. void (*pfunc)(void*); declare pfunc pointer function return type void , expects argument of type void *.

why output: aa b ee f ii j?

the snippet

for(i=1; i< argc;i++)        pfunc(argv+i);   

passes of strings aaa, eee , iii function print. there casting a char **.
first string aaa, *p pointer first char , *p+1 second char. statement

printf("%s",(*p+1));   

will print string second character, i.e print aa.
**p char , **p+1 increment ascii value of character , putchar(**p+1); print character. ascii value of a 97 , 98 b.

therefore, output aaa aa b. same goes other arguments.