C argv[2]++ vs (char*)argv[2]++?

Question

Why are argvs act odd? Example:

This will work as expected, print the 1st character of the 1st argument.

printf("%c", *argv[1]);

This however will print the character that's in the ascii table (aka "one bigger" represented as a number) instead of printing the 2nd character of the 1st argument:

printf("%c", *(argv[1] + 1);

Slight modifications will do nothing, exact same result:

printf("%c", *(argv[1] + sizeof(char));

The only thing that helps is actually casting argv[2] to char*:

printf("%c", *( (char*)(argv[1] +1 ) );

Why is this happening? I kno that:

argv's type :=  char** argv
and argv[1] == *(argv + 1)
and argv[1]'s type should be char*

but apparently it is not, as I need to cast it? How is this possible? It is declared as char**, yet its members aren't char*s?

If you want to index characters, why not `argv[1][0]`, `argv[1][1]`, etc.? — crashmstr, Sep 23 '15 at 19:54
You have a missing `)` in the second snippet. Perhaps if you fix it, it will work as expected. Update: And the third. And fourth. — Eugene Sh., Sep 23 '15 at 19:55
The output you claim to get would be from `*argv[1]+1, not `*(argv[1]+1)`. — Barmar, Sep 23 '15 at 20:13

score 3 · Answer 1 · answered Sep 23 '15 at 19:55

3

The premise of your question is mistaken. This program:

#include <stdio.h>

int main (int argc, char *argv[]) {
  printf("%c\n", *(argv[1] + 1));
  return 0;
}

run with the argument test, prints e, not u.

answered Sep 23 '15 at 19:55

hobbs

223,387
19
210
288

I'll defer to your 6 years of experience here, but I didn't think that pointing out that the question was mistaken is usually an answer. – Barmar Sep 23 '15 at 20:15
@Barmar it's not, but you can't do code formatting in a comment. I don't care if it gets deleted as long as OP reads it first :-P – hobbs Sep 23 '15 at 20:17

John Bode · Answer 2 · 2015-09-23T20:39:56.173

0

Make sure your parentheses and dereferences are really where you think they are. If argv[1] contains test, then

* ( argv[1] + 1 ) == 'e'
( * argv[1] + 1 ) == 'u'
^ ^
| |
+-+- slight difference here

edited Sep 23 '15 at 20:39

answered Sep 23 '15 at 20:28

John Bode

119,563
19
122
198

C argv[2]++ vs (char*)argv[2]++?

2 Answers2