(int) casting

[Azmeos]

When I try to typecast a user-input value, I'm getting the address (I believe) instead of the actual value. Can someone help?

#include <iostream>

using namespace std;

int main(int argc, char *argv[]) {
	if (argc < 1) {
		cout << "ERROR: You must supply a number." << endl;
		return 0;
	}

	int number = (int)argv[1];

//more code after this...
}

[Azmeos]

Maybe it's easier to ask how I get an INT from the user input. Even a CIN would work... help?

[Azmeos]

I figured out CIN ... but I'm still wondering how I get the number from the command-line arguments. Anyone know?

[btq]

argv[1] is a char-string, not a number.
To convert it to an int look up atoi().

/btq

[hk_mp5kpdw]

if (argc < 1) {

Should be:

if (argc < 2) {

Otherwise simply running the program without an argument (argc will always equal at least 1) will attempt to access an invalid memory location, i.e. argv[1]. As you have it, the ERROR message will never get output.

[JaWiB]

don't you have to dereference it?

	int number = (int)*argv[1];

//more code after this...

[laasunde]

atoi

[XSquared]

#include <iostream>
#include <cstdio>

using namespace std;
int main(int argc, char *argv[]) {
  
  if (argc < 2) {

    cout << "ERROR: You must supply a number." << endl;
    return 0;

  }
  
  int number;
  if( sscanf( argv[ 1 ], "%d", &number ) == 1 ) {
    
    //number now holds the integer value of argv[ 1 ]
    
  } else {
    
    //there was an error
    
  }
  
//more code after this...
}