If argc is 1, can I still use argv[1], argv[2],... character arrays?

Question

If there is no argument passed from the command line ie. if argc is 1, can we still allocate memory for argv[1],argv[2],..... and use those buffers for further experiments. If that is undefined behavior, can I still use it somehow?

If `argc` is `1` then accessing `argv[1]` or higher is probably UB. — Joachim Sauer, Sep 26 '22 at 10:15
@JoachimSauer If that leads to UB, can I somehow use it? workaround? — Saurav Rai, Sep 26 '22 at 10:18
@JoachimSauer If `argc` is `1` then `argv[1]` will be accessible, but it will contain a null pointer. — Ian Abbott, Sep 26 '22 at 10:19
@IanAbbott what about {argv[2], argv[3],...} Is it the same case? can we use it? — Saurav Rai, Sep 26 '22 at 10:20
`argv[argc] == NULL` (not 100% sure that is C standard, it might also be just POSIX or implementation specific). Accessing `argv[argc+1]` or beyond is definitely Undefined Behavior. You may modify pointers in valid range of `argv`, and it is actually somewhat common practice (to hide command line arguments which are already handled, keeping unhandled ones available for other code). — hyde, Sep 26 '22 at 10:20
@IanAbbott: fair, my C days are long ago. I think what I said still applies to `argv[2]` and up though. — Joachim Sauer, Sep 26 '22 at 10:21
@hyde `argv[argc]` is required by the C Standard to be `NULL`. [cppreference.com](https://en.cppreference.com/w/c/language/main_function) — Dúthomhas, Sep 26 '22 at 10:27
It is more like if your cars is announced to have 2 seats, can you use seat #3 and up? No, you cannot, and you cannot "make" more seats, as the car is immutable. — the busybee, Sep 26 '22 at 10:27
It is also possible for `argc` to be `0`, in which case `argv[0]` will be `NULL`. — Ian Abbott, Sep 26 '22 at 10:29
See also [Is it possible to add an argument to main by scanf?](https://stackoverflow.com/q/30904917/2410359). — chux - Reinstate Monica, Sep 26 '22 at 11:50

score 3 · Answer 1 · answered Sep 26 '22 at 10:26

No, the C standard does not specify that argv has any elements beyond argv[argc], so they may not exist in C’s object-memory model and the behavior of using them is not defined by the C standard.

C 2018 5.1.2.2.1 2 says:

…

argv[argc] shall be a null pointer.

If the value of argc is greater than zero, the array members argv[0] through argv[argc-1] inclusive shall contain pointers to strings, which are given implementation-defined values by the host environment prior to program startup.

…

That is all there is that defines the extent of the argv array; nothing in the standard says there are more elements.

When argc is one, using argv[1] is defined but using argv[2] is not.

You can store new values to the defined elements because C 2018 5.1.2.2.1 2 also says:

The parameters argc and argv and the strings pointed to by the argv array shall be modifiable by the program, and retain their last-stored values between program startup and program termination.

score 1 · Answer 2 · answered Sep 26 '22 at 10:30

1

can I access argv[] elements after argv[argc]?

You can ... but ONLY IN THIS CODE

#include <stdio.h>

int main(int argc, char **argv) {
    if (argc == 1) {
        char *foo[] = {"bar", "baz", "quux", NULL, "bingo"};
        main(3, foo);
    } else {
        printf("argc is %d; argv[4] is \"%s\"\n", argc, argv[4]);
    }
    return 0;
}

See code running on ideone.

In all other codes, you cannot.

answered Sep 26 '22 at 10:30

pmg

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Interesting !!. – Saurav Rai Sep 26 '22 at 10:36
That only works if the original `argc` satisfies `argc == 1 || argc > 4`. – Ian Abbott Sep 26 '22 at 10:37
Yeah, snippet can be made more robust @IanAbbott :) – pmg Sep 26 '22 at 10:39
@pmg So if I have to use dynamic memory allocation instead of this static allocation. I need to free the character buffers...hence this won't work – Saurav Rai Sep 26 '22 at 10:50
1

Keep it simple @SauravRai, don't try and abuse `argc` and `argv`. – pmg Sep 26 '22 at 10:55

score 0 · Answer 3 · answered Sep 26 '22 at 10:22

0

Always argv[argc] is equal to NULL. In the described case where argc is equal to 1 the array argv contains two pointers argv[0] and argv[1] where argv[1] is a null pointer.

You may reassign the pointers but this does not make a great sense because that will make your program unclear. Instead you could declare your own array if you need.

answered Sep 26 '22 at 10:22

Vlad from Moscow

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"You may reassign the pointers" --> [Is `argv[n]` writable?](https://stackoverflow.com/q/25737434/2410359) – chux - Reinstate Monica Sep 26 '22 at 11:49
@chux-ReinstateMonica argv is not an array of constant pointers. So the elements of the array may be reassigned. – Vlad from Moscow Sep 26 '22 at 11:58
C spec treats `main()` special. Also, lack of `const` does not mean changing is OK. E.g. `char *s = "abc";`. IAC, that old [question](https://stackoverflow.com/q/25737434/2410359) digs into the matter and your insight is appreciated there. – chux - Reinstate Monica Sep 26 '22 at 12:02
@chux-ReinstateMonica Elements of the array are pointers. So to write for example argv[1] = "abc"; is a valid code. main is special in the meaning of how the array argv is initialized and nothing more. – Vlad from Moscow Sep 26 '22 at 12:06

If argc is 1, can I still use argv[1], argv[2],... character arrays?

3 Answers3