How can I succesfully call the execv function?

Question

I am trying to make a program that will copy file1 into file2 the following way:

cp -i -p file1 file2

Now I call my executable copy and so by calling

copy file1 file2

It will do the same thing like the first command (-i and -p).

I was able to do this using execl

char const *copy[] = {"/bin/cp","cp","-p","-i",0};

execl(copy[0],copy[1],copy[2],copy[3],argv[1],argv[2],copy[4]);

However, I want to do it now with execv

I saw the man page of exec* functions

execl(const char *path, const char *arg, ...);

execv(const char *path, char *const argv[]);

and so the first argument seems to be the same however,

How the second argument for execv is char *const argv[]

what do I need to change in the execv function to get the same result ?

I have my main function arguments like the following:

main(int argc,char * argv[])

Solution

Change your copy array, and the function call. The following is a minimal example:

#include <unistd.h>

int main(int arcg, char *argv[])
{
    char *const args[] = {"cp","-p","-i", argv[1], argv[2], 0}; 
    execv("/bin/cp", args);
}