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I think my program handles lack of argv[1] correctly, yet it fails check50's test. here my code:

if ( argc != 2 ) printf ("usage : ./caesar + key \n"); return 1;

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4 Answers 4

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the issue is that you started your code

int main( int argc, string argv[])

{ int k = atoi(argv[1];
. . . }

/// the mistakes is in the above assignment, if the there is no argv[1] so the OS will report error and terminates the program

so you have not to do k assignment to argv[1] unless you make sure there is

argv[1]

thanks

Mohamed Abdeltawab

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  • This solved the issue I was having! thank you! Commented Feb 14, 2017 at 16:51
  • Thanks! That helped me 🙌 Commented Apr 8, 2017 at 19:29
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The number of command-line arguments cannot be negative. In fact, it cannot even be < 1. See this answer for more details!

If the number of command-line arguments is not 1 (i.e., argc != 2), your program should "print" an error message and "return" an error code of 1.

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  • I assume that it is sufficient to test for argc != 2. But when I implemented this, and had the program indeed issue the exit(1) when argc != 2 was True, I got the following. :( handles lack of argv[1] \ expected output, not an exit code of 1 I'm confused, as I believe I'm doing what is expected of me by the program requirements - that is, issuing the exit(1). What am I misinterpreting? Thanks. Commented Oct 2, 2014 at 1:54
  • I've also used the following condition in doing the check - if ((argc<2) || (argc>2)) -with the same result. Commented Oct 2, 2014 at 2:10
  • @rl777 please read the answer carefully! "your program should "print" an error message and "return" an error code of 1" Commented Nov 28, 2014 at 20:01
  • Where does it say in the pset specification to print something? Did yell imply so? Commented Dec 11, 2014 at 18:08
  • Yes, @maq...... Commented Dec 11, 2014 at 18:10
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You need to add your curly brackets (aka braces). Instead of 

if ( argc != 2 ) printf ("usage : ./caesar + key \n"); return 1;

You should have:

if ( argc != 2 ) { printf ("usage : ./caesar + key \n"); return 1; }
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  • It's better to use, printf ("usage : ./caesar + k \n"); Because the instruction code also gives such examples. Commented Jan 11, 2018 at 19:33
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if (argv[1] == NULL) { printf("Usage: ./caesar key\n"); return(1); } this is the solution before the program received the first statement validate null

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  • Thanks so much, solved since 2014 But you are very kind and I went back to stack Commented Mar 1, 2021 at 11:57

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