sizeof argv[1] not working

Question

I'm really new to C and all I know is that the error is related to oldname and newname not be initialized

#include <stdio.h>

int main (int argc, char const *argv[])
{
    int result;
    int lengthOne;
    int lengthTwo;
    lengthOne = sizeof(argv[0]);
    lengthTwo= sizeof(argv[1]);

    char oldname[lengthOne] = argv[0];
    char newname[lengthOne] = argv[1];

    result = rename(oldname, newname);

    if (result == 0) {
        puts "File renamed";
    } else {
        perror "ERROR: Could not rename file";
    }

    return 0;
}

app.c: In function ‘main’:
app.c:11: error: variable-sized object may not be initialized
app.c:12: error: variable-sized object may not be initialized
app.c:17: error: expected ‘;’ before string constant
app.c:19: error: expected ‘;’ before string constant

Answer 1

lengthOne = sizeof(argv[0]);
lengthTwo= sizeof(argv[1]);

This gives you the size of a char*, not the length of the string. You meant strlen, not sizeof.

char oldname[lengthOne] = argv[0];
char newname[lengthOne] = argv[1];

You can't assign to arrays like that. You could strcpy them, but it's unnecessary here because you can just use pointers.

const char* oldname = argv[0];
const char* newname = argv[1]; // But verify that argc >= 2 first!

EDIT: Also, don't forget that argv[0] is the name of the program itself, and argv[1] is the first argument. If your intent was to write a mv-like program instead of a program that renames itself, you want argv[1] and argv[2].