Answer by Nobilis for cannot convert parameter 1 from 'char [20][20]' to...
char** is a pointer to a pointer.You want a pointer to an array (hint: it's not char* [] either as this is an array of pointers equivalent to the above).You need char (*)[size] (notice the brackets)....
View ArticleAnswer by Mike Woolf for cannot convert parameter 1 from 'char [20][20]' to...
Yeah, this is where the idea that an array can be treated like a pointer falls apart."char[20][20]" denotes an array of 400 characters, laid out in a 20x20 fashion. It is not an array of 20 pointers to...
View Articlecannot convert parameter 1 from 'char [20][20]' to 'char ** '？
My Code is below:#include <stdio.h>void print_pointer(char **str);void print_array(char *str[20]);void print_array2(char str[20][20]);void print_array3(char str[][20]);int main(int argc, char...
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