pass string into main

hi
i need to pass a string into a program of mine but this doesn't seem to be working.

its been a while but this isnt working and im frustrated :(

Code:

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#include "stdio.h"
#include "string.h"

void main(char badsmell[30]){
        printf("%s", badsmell);
        return;
}

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Read the FAQ (or any good book) to find out about argc and argv.

also

Code:

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#include <cstdio>
#include <cstring>

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Code:

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#include <string.h>

int main(char** argv, int argc)
{

    char badsmell[30];

    if (argc == 1)
    {

        strcpy(badsmell, argv[1]);

    }

}

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Code:

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#include <string.h>

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That's <cstring> in C++.

Code:

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int main(char** argv, int argc)

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That's the other way round.

Code:

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    char badsmell[30];

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How do you know this is long enough?

Code:

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    if (argc == 1)
    {

        strcpy(badsmell, argv[1]);

    }

---

This will always go out of bounds.

Seriously, watch Salem's avatar until you understand the message it contains.

Code:

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#include <cstdio>
#include <cstdlib>
#include <cstring>

int main(int argc, char **argv)
{
        char *badsmell = NULL;
       
        if (argc == 2)
        {
                int size = strlen(argv[1]);
                badsmell = (char *)malloc(sizeof(char) * size);
                strcpy(badsmell, argv[1]);
               
                printf("&#37;s\n", badsmell);
        }
       
        if (badsmell) free(badsmell);
        return 0;
}

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Of course this being C++, you could also do

Code:

---

#include <iostream>
#include <string>
using namespace std;

int main(int argc, char **argv)
{
        string badsmell;
       
        if (argc == 2)
        {
                badsmell = argv[1];
               
                cout << badsmell << endl;
        }
       
        return 0;
}

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Quote:

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Originally Posted by iMalc

Seriously, watch Salem's avatar until you understand the message it contains.

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That's a good point. Never use void main().