Pass Array of String by Reference in c

Question

I want to proceed an Array of string Arg[10][50] using function strtoArg(), It should be return two values one is variable 'ArgNo' which is Numbers of Arg (would pass by value) & other is Arguments array variable 'arg' (would pass by reference). Help me to solve this problem, how can I get proper values of Arg return back from function.
Thanks in advance.

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
#define IN 
#define OUT


int strtoArg (IN char* instr, OUT char &arg)
{

char *pch,temp[10][50];
    int ArgNo=0,j;  

    pch = strtok (instr," -");
    while (pch != NULL){
        strcpy((char*)(temp[ArgNo]),pch);
        arg = &temp[ArgNo]; 
        pch = strtok (NULL, " -");
        ArgNo++;    
    }
    return ArgNo;
}


int main()

{

    char Instr[50],Arg[10][50];
    int ArgNum,i;

    gets(Instr);

    ArgNum = strtoArg(Instr,*Arg);

    for(i=1; i<ArgNum; ++i)
        printf("%s",Arg[i]);        

    return 0;
}

I have try to go this way,

int strtoArg (IN char* instr, OUT char* arg)
{    
    char *pch;
    int ArgNo=0,j;  

    pch = strtok (instr," -");
    while (pch != NULL){
        strcpy((char*)arg[ArgNo],pch);
        pch = strtok (NULL, " -");
        ArgNo++;    
    }
    return ArgNo;    
}

int main()
{       
    char Instr[50],Arg[10][50];     
    int ArgNum,i;    
    gets(Instr);

    ArgNum = strtoArg(Instr,*Arg);

    printf("here"); 

    for(i=1; i<ArgNum; ++i)
        printf("%s",Arg[i]);        

    return 0;
}

It's remove error but create Segmentation fault. Am I on right path?

there's a C++ism... should tag c++?

ShinTakezou
– ShinTakezou

2011-07-20 07:14:16 +00:00
Commented Jul 20, 2011 at 7:14 — ShinTakezou
– ShinTakezou, Commented Jul 20, 2011 at 7:14

Stan · Accepted Answer · 2011-07-20 07:45:55Z

1

I revised your code, see following:

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
#define IN
#define OUT


int strtoArg (IN char* instr, OUT char arg[][50])
{
    char *pch;
    int ArgNo=0,j;

    pch = strtok (instr," -");
    while (pch != NULL){
        strcpy(arg,pch);
        arg++;
        pch = strtok (NULL, " -");
        ArgNo++;
    }
    return ArgNo;
}


int main()
{

    char Instr[50],Arg[10][50];
    int ArgNum,i;

    gets(Instr);

    ArgNum = strtoArg(Instr, Arg);

    for(i=1; i<ArgNum; ++i)
        printf("%s\n",Arg[i]);

    return 0;
}

Pitfalls in your code are:

ArgNum = strtoArg(Instr,*Arg); in main(). That *Arg doesn't mean the address of the 2D array.
int strtoArg (IN char* instr, OUT char &arg). &arg is a reference in C++, it means you will modify the variable in the caller when calling the function. However, what you pass is just an array name, and it's not necessary to change value for itself.
arg = &temp[ArgNo]; seems you want to "return" a local array, unfortunately, you shouldn't do this. You cannot return any local data that stores in the stack, because it would be broken after the function returning.

Aside: the parameter char arg[][50], arg means 'a pointer to a char[50] array', when arg++ it goes to 'the next row' of Arg[10][50]. Hmmm, although it may fit your need, maybe, it lacks flexibility.

edited Jul 20, 2011 at 7:45

answered Jul 20, 2011 at 7:28

Stan

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3 Comments

Hit's Over a year ago

Thank you very much Stan your solution code works!!!!! But one thing I have to change is, in stand of 'strcpy(arg,pch);' it should be 'strcpy(*arg,pch);' after this it works perfect!!!!!!!!! Can you please give reason of it?

Stan Over a year ago

@Heet Kansagra: My code works well on my computer. In fact, the function can be written as: int strtoArg (IN char* instr, OUT char (*arg)[50]), that is to say, arg can be considered as a pointer to char[50] array. However, in C, the address of array has chaotic meanings, you can try:

int main(){ char Arg[3][2]={1,2,3,4,5,6}; char (*p)[50] = Arg; printf("p=%x\n", p+1); printf("*p=%x\n", *(p+1)); return 0;}

. Amazingly, the output will be the same. Briefly speaking, &Arg and Arg are equal in value (although their types are different).

Hit's Over a year ago

as you explain arg should be an pointer at the time of function declaration or when we copy string in it. Yes, Arg and &Arg are same but only if we declarer it as a pointer. By the way Thanks Again...!!

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