pass string into main

[chico1st]

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

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

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

[Salem]

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

[robwhit]

also

#include <cstdio>
#include <cstring>

[francoissoft]

#include <string.h>

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

    char badsmell[30];

    if (argc == 1)
    {

        strcpy(badsmell, argv[1]);

    }

}

[anon]

#include <string.h>

That's <cstring> in C++.

int main(char** argv, int argc)

That's the other way round.

    char badsmell[30];

How do you know this is long enough?

    if (argc == 1)
    {

        strcpy(badsmell, argv[1]);

    }

This will always go out of bounds.

[iMalc]

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

[Frobozz]

#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;
}

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

#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;
}

[francoissoft]

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

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