a pointer about *argv[]

Question

This is my main.c

......
int main(int argc, char **argv)
{
    init_arg(&argc, &argv);
    ......
}

This is my init_arg.c

......
void init_arg(int *argc, char ***argv)
{
    printf("%s\n", *argv[1]);
    ......
}

I compiler it with no error and warning.

I run it:

./a.out include

It get Segmentation fault

When I debug it， I found step printf("%s\n", *argv[1]);

get wrong, It show:

print *argv[1]

Cannot access memory at address 0x300402bfd

I want to know, How to print argv[1] in init_arg() function.

Answer 1

You need to add a pair of parentheses around (*argv) to change the order of evaluation. The way you currently have it, the [1] is evaluated first, yielding an invalid pointer, which then gets dereferenced, causing undefined behavior.

printf("%s\n", (*argv)[1]);

Answer 2

Argv is already a pointer. Just pass it like this:

init_arg(&argc, argv);

And init_arg should look like this:

void init_arg(int *argc, char **argv) {
    printf("%s\n", argv[1]);
}