Program stops working after printing argv

Question

I wrote some basic code

#include <iostream>
using namespace std;

int main( int argc, char *argv[] ){
    for (int i=1;i<=argc;i++){
        cout << argv[i] <<"\n";
    }
    cout <<"hello";

    return 0;
}

and when i'm running it for example with

./a.out 1 2 3

I see:

1
2
3

Why there's no "hello"?

Probably because you're shy a `flush`. – Rubens Mar 29 '14 at 21:09 — Rubens, Mar 29 '14 at 21:09
replace `i<=argc` with `i – Iłya Bursov Mar 29 '14 at 21:10 — Iłya Bursov, Mar 29 '14 at 21:10
@Rubens: No, streams are flushed on program end. – Lightness Races in Orbit Mar 29 '14 at 21:14 — Lightness Races in Orbit, Mar 29 '14 at 21:14

score 4 · Accepted Answer · edited May 23 '17 at 12:20

4

Your loop tries to stream argv[argc], which is one after your program arguments.

In fact, it is defined to be a null pointer¹, and giving streams a null pointer sets their error bit².

Consequently, your next stream operation (cout << "hello") fails.

Loop up to argc but not including it:

for (int i = 1; i < argc; i++) {
    cout << argv[i] << '\n';
}

Footnote 1

[C++11: 3.6.1/2]: [..] The value of argv[argc] shall be 0. [..]

Footnote 2

[C++11: 27.7.3.6.4/3]: Requires: s shall not be a null pointer

When it is, the behaviour is undefined; GCC chooses to trap the condition and set the stream's error bit — see https://stackoverflow.com/a/7019483/560648.

edited May 23 '17 at 12:20

Community

1
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answered Mar 29 '14 at 21:20

Lightness Races in Orbit

378,754
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1,055

and, of course, argv[0] is the name of the executable. which is why the first argument is at an un-C-like "1". – Roddy Mar 29 '14 at 21:26
@JohannesSchaub-litb: Are you joking? – Lightness Races in Orbit Mar 29 '14 at 21:41
@LightnessRacesinOrbit no, definitely not – Johannes Schaub - litb Mar 29 '14 at 21:57
@JohannesSchaub-litb: I'm sorry to hear that :( – Lightness Races in Orbit Mar 29 '14 at 22:20

score 1 · Answer 2 · answered Mar 29 '14 at 21:11

1

you're referencing argv[argc] which is beyond the array.

Change your for statement to:

for (int i=1;i<argc;i++){

answered Mar 29 '14 at 21:11

zumalifeguard

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Program stops working after printing argv

2 Answers2

Footnote 1

Footnote 2