Issue

Below is the code of main() of grub. Here I want to know about this line:

file = fopen(arg_v[1], "rb");

Here which file fopen is opening? What file this arg v[1] is pointing to?

int main(unsigned arg_c, char *arg_v[])    
{    
    FILE *file;    

    if(arg_c < 2)    
    {    
        printf("Checks if file is Multiboot compatible\n");    
        return 1;    
    }    
    file = fopen(arg_v[1], "rb");    
    if(file == NULL)    
    {    
        printf("Can't open file '%s'\n", arg_v[1]);    
        return 2;    
    }    
    check_multiboot(arg_v[1], file);    
    fclose(file);    
    return 0;    
}

Solution

If you call your program with

program arg1 arg2.txt 65

argv[1] is a pointer to "arg1"; argv[2] is a pointer to "arg2.txt", argv[3] is a pointer to "65", argv[4] is NULL

argv[0] either points to "program" or to "" if the OS and/or Library and/or startup code cannot identify the name used to call the binary executable

In your specific case, the program tries to open a file, whose name is provided in the first argument to the program, in read binary mode.

Answered By - pmg

Answer Checked By - Robin (WPSolving Admin)

Thursday, March 17, 2022

[SOLVED] Query regarding main() of GRUB

Issue

Solution