Error in copying argv to char array

Question

I am trying to copy argv to char array, been through some solutions online but ending up in getting Segmentation Fault. Following is the code i used:

void main (int argc,const char *argv[])
{
    char *arr;
    arr = (char *) malloc(strlen(argv[1])+1);
    strcpy(arr,argv[1]);
}

Please help to identify what I am doing wrong.

Which parameter did you specify when running the program? You should check for `argc > 1` before allocating, copying. — πάντα ῥεῖ, Oct 31 '14 at 08:01
what is `argv[1]` ? `strlen(NULL)` will lead to UB and segfault. See [this](http://stackoverflow.com/q/5796103/2173917) — Sourav Ghosh, Oct 31 '14 at 08:02
And you pass at least one argument to the program? Please tell us how you invoke the program and what arguments you pass to it. — Some programmer dude, Oct 31 '14 at 08:04

Vlad from Moscow · Accepted Answer · 2014-10-31T08:22:26.370

5

It seems that argv[1] is equal to NULL or even does not exist (The C Standard allows that argc may be equal to 0).

Add the following check

char *arr;

if ( argc > 1 )
{
    arr = (char *) malloc(strlen(argv[1])+1);
    strcpy(arr,argv[1]);
}
else
{
    // print some error message
}

edited Oct 31 '14 at 08:22

answered Oct 31 '14 at 08:03

Vlad from Moscow

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@Sourav Ghosh I have not understood why you downvoted my answer. – Vlad from Moscow Oct 31 '14 at 08:08
I just added some more info. See my previous comment. isn't it same as urs? why i'll downvote then? – Sourav Ghosh Oct 31 '14 at 08:13
@Sourav Ghosh I do not know but my answer is downvoted. So it is interesting to know why it is downvoted. – Vlad from Moscow Oct 31 '14 at 08:14
ditto. I also expect a comment stating the reson for a downvote. – Sourav Ghosh Oct 31 '14 at 08:16

4pie0 · Answer 2 · 2014-10-31T08:20:53.780

Please help to identify what I am doing wrong.

All right then sir. You are asking for argv[1], but you are not sure whether it exists. accessing an array outside its bounds has undefined behavior. You should always check if number of parameters is what you expect to avoid undefined behavior:

if ( argc < 2 )
{
    // error, cannot copy argv[1] because it doesn't exist. Explain this to user
}

// now OK..., also we postponed allocation of arr pointer 
char *arr =  malloc( strlen( argv[1]) + 1);
        //^^^^
        // no need to cast return value of malloc in C

strcpy( arr, argv[1]);

score 0 · Answer 3 · edited Oct 31 '14 at 09:54

When using command line input, we should deal with number of arguments.

You can try something like this..

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

void main (int argc, const char *argv[])
{
 if(argc==2)//change condition based on your requirements
 {
  char *arr;
  arr = (char *) malloc(strlen(argv[1])+1);
  strcpy(arr,argv[1]);
  printf("string is %s\n",arr);
 }
else
 {
 printf("check your command line input (only 2 parameters)\n");
 }
}

OUTPUT:

 $ ./a.out 
 check your command line input (only 2 parameters)
 $ ./a.out hello
 string is  hello
 $ ./a.out hi hello
 check your command line input (only 2 parameters)
 $

Error in copying argv to char array

3 Answers3