CODEWITHSUNDEEP
c++program-entry-point
Little
Little

Reputation: 3477

Arguments for main() in C++

I have been following the book Flex & Bison from Levine and I see the following piece of code:

main(argc, argv)
int argc;
char **argv;
{
    if(argc > 1) {
        if(!(yyin = fopen(argv[1], "r"))) {
            perror(argv[1]);
            return (1);
        }
    }
    yylex();

I find it little hard to understand this part, for example, the first line:

char **argv;

for what I can see it should be related to this part:

if(!(yyin = fopen(argv[1], "r")))

so I can interpret that it is reading an input operation, this because of the "r", and the argv[1] refers to the input of a line by console. Is it like that? I have no clue why the author is using a double pointer to argv. I suppose that when there is not a valid input it prints a descriptive error with perror(), but in which case this could happen?

Upvotes: 1

Views: 118

Answers (1)

RealPawPaw
RealPawPaw

Reputation: 986

The char **argv is interpreted by the C++ compiler as the same as char* argv[], i.e, a pointer to an array of pointers. As you may already know, argv represents the arguments passed to the program executable, e.g. ./program arg1 arg2. You can think of argv as an array of strings (C-strings) if that simplifies things, each element of the array is an argument.

Your interpretation is basically right. It checks that there is at least one argument argc > 0 (argument count), then uses the first argument (argv[1] as argv[0] actually refers to the program name ./program) as a file name. It then tries to open a file of name argv[1] for reading. If this fails, the fopenreturns NULL. This can happen if the file doesn't exist.

Upvotes: 1

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