Help with command-line prog

[McCarthy]

Im writing a command-line program in unix. Whenever I run the prog in unix it gives me 49 output files instead of the user entered amount. Im not sure what the problem is with my main prog the loop looks fine

 
int main(int argc, char * argv[])
{
    // Command Arguements
 
    // Usage: <program> <policy file> <gen file prefix> <num generations>
    // argv[1] is the name of the policy file
    // argv[2] is the string that the generation output file starts with 
    // argv[3] the number of generations that will be evolved in this simulation
 
       if(argc != 4) {
        cout << "Usage:\n   " << argv[0] << " <policy file> <gen file prefix> <num generations>" << endl;
        return -1;
    }
 
    Policy thePolicy(argv[1]);
    Generation * gPtr[*(argv[3])];
    gPtr[0]= new Generation(thePolicy, argv[2]);
    for (int i=1; i<=(*(argv[3])); i++)
          gPtr[i]= new Generation(thePolicy, gPtr[i-1], argv[2]);
    return 0;

[weaknessforcats]

This code:

Generation * gPtr[*(argv[3])];
 

Isn't argv[3] a char*? As in a C string?

If the argv[3] string is 10, then argv[3] is the address of the 1. Therefore,
*argv[3] will be the character 1, which has an integer vaue of 49.

[McCarthy]

This code:

Generation * gPtr[*(argv[3])];
 

Isn't argv[3] a char*? As in a C string?

If the argv[3] string is 10, then argv[3] is the address of the 1. Therefore,
*argv[3] will be the character 1, which has an integer vaue of 49.

So how would I go about making the number of generations be user defined?

[McCarthy]

This code:

Generation * gPtr[*(argv[3])];
 

Isn't argv[3] a char*? As in a C string?

If the argv[3] string is 10, then argv[3] is the address of the 1. Therefore,
*argv[3] will be the character 1, which has an integer vaue of 49.

Yea thanx I got it I just have to use atoi(argv[3]). I didnt realize that it was a character*.