strncpy from **char to *char[] not copying last element

Question

I'm trying to copy the individual elements of argv into a 2nd array. I'm using strncpy to do it.

Here's a stripped version that replicates the problem.

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

int main(int argc, char *argv[])
{
    printf("%d\n", argc);
    char *argvarray[argc];
    int arrlen;

    //int j;
    for(int j=0; j< argc; j++){
        arrlen = arrlen + strlen(argv[j]);
    }

    printf("size of argv: %d\n", arrlen); //see if size is right
    strncpy(*argvarray, *argv, arrlen);

    for(int j=0; j< argc; j++){
        printf("value of j: %d\n", j); //to see what's going on
        printf("%s\n", argv[j]); 
        printf("%s\n", argvarray[j]); //gets 1st element no problem
    }

    return 0;
}

If I pass 1 arg to main it copies the first element of argv to argvarray but the second is not copied. If I pass more than 1 arg it segfaults on strncpy().

Why is this happening? What's the best way to copy all elements of a char array (particularly char arrays declared in function definitions) to other char arrays without knowing the size before hand, as in the case of argv?

You never allocate memory to which the elements in `argvarray` could point to. — Peter - Reinstate Monica, Feb 19 '16 at 13:40
Thanks guys, I appreciate the assistance, and your patience as I learn. — Jack Galt, Feb 19 '16 at 14:20
Study arrays and pointers in detail before moving on to string handling. — Lundin, Feb 19 '16 at 14:22
@Lundin That's exactly what I'm doing, studying arrays and pointers. I'm a 30 year programmer (of a different sort) and I suspect that's actually a detriment to me learning C. But, I learn best through practical use of individual concepts as opposed to volumes of discourse on theory and arbitrary example. Since I primarily work with strings and ints it's imperative that I understand how to apply arrays and pointers to strings and ints. Ergo, the request for assistance in learning how to apply them. — Jack Galt, Feb 19 '16 at 14:43

David Ranieri · Answer 1 · 2016-02-19T13:45:18.260

3

arrlen = arrlen + strlen(argv[j]);

arrlen is used uninitialized (undefined behaviour).

And as pointed out By @PeterA.Schneider, you need room for each element of argvarray, use malloc before strncpy or strdup

edited Feb 19 '16 at 13:45

answered Feb 19 '16 at 13:40

David Ranieri

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Thanks I appreciate all the constructive feedback. – Jack Galt Feb 19 '16 at 14:28

score 1 · Answer 2 · answered Feb 19 '16 at 13:43

1

You're confused about arrays vs pointers.

If you declare an array char tab[2][3], it'll be stored contiguously in memory:

[h][e][y][y][o][u] (a simple example).

But here, you have pointers, so char (*tab)[2] is an array of two pointers:

tab = [0x1234][0x5678]
0x1234 = "hey"
0x5678 = "you"

strncpy copies contiguous parts of memory. To copy argv, copy each pointer instead (call malloc + strcpy one for each char* is the easiest option)

(also see @AlterMann's answer, which is correct as well)

answered Feb 19 '16 at 13:43

Ven

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YES! Arrays are kicking my butt. Pointers aren't much better. If I weren't in need of a better understanding of C I'd probably rather spend time juggling knives. I'll get there eventually though (I hope) – Jack Galt Feb 19 '16 at 14:31

score 1 · Answer 3 · answered Feb 19 '16 at 13:44

arrlen is not initialised. You increase its value then attempt to use it as a parameter to strncpy. The program behaviour will be undefined. Write int arrlen = 0; instead.
argvarray is an array of pointers to char. You don't allocate any memory for each element. *argvarray is equivalent to argvarray[0] which is a char* that doesn't point to anything. So, the behaviour of strncpy will be undefined.

score 0 · Answer 4 · answered Feb 19 '16 at 13:52

Apart from that variable arrlen was not initialized

int arrlen;

and as result the value of the variable after execution the loop

for(int j=0; j< argc; j++){
    arrlen = arrlen + strlen(argv[j]);
}

will be indeterminate this call

strncpy(*argvarray, *argv, arrlen);

does not make sense and has undefined behaviour.

I think what you mean is the following

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

int main(int argc, char *argv[])
{
    printf("%d\n", argc);
    char *argvarray[argc];

    for ( int i = 0; i < argc; i++ )
    {
        argvarray[i] = malloc( strlen( argv[i] ) + 1 ); 
        strcpy( argvarray[i], argv[i] );
    }

    for ( int i = 0; i < argc; i++ )
    {
        printf( "value of i: %d\n", i ); //to see what's going on
        printf( "%s\n", argv[i] ); 
        printf( "%s\n", argvarray[i] ); //gets 1st element no problem
    }

    for ( int i = 0; i < argc; i++ ) free( argvarray[i] );

    return 0;
}

Thanks a million. I see how the other answers were basically the same as yours, but your example of usage really helped clarify it. Quality answer!! — Jack Galt, Feb 19 '16 at 14:17

strncpy from **char to *char[] not copying last element

4 Answers4