(int) casting

This is a discussion on (int) casting within the C++ Programming forums, part of the General Programming Boards category; When I try to typecast a user-input value, I'm getting the address (I believe) instead of the actual value. Can ...

  1. #1
    Registered User Azmeos's Avatar
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    (int) casting

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

    Code:
    #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...
    }

  2. #2
    Registered User Azmeos's Avatar
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    Maybe it's easier to ask how I get an INT from the user input. Even a CIN would work... help?

  3. #3
    Registered User Azmeos's Avatar
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    I figured out CIN ... but I'm still wondering how I get the number from the command-line arguments. Anyone know?

  4. #4
    btq
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    julie lexx... btq's Avatar
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    argv[1] is a char-string, not a number.
    To convert it to an int look up atoi().

    /btq
    ...viewlexx - julie lexx

  5. #5
    Registered User hk_mp5kpdw's Avatar
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    Code:
    if (argc < 1) {
    Should be:
    Code:
    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.
    "Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
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  6. #6
    carry on JaWiB's Avatar
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    don't you have to dereference it?
    Code:
    	int number = (int)*argv[1];
    
    //more code after this...
    "Think not but that I know these things; or think
    I know them not: not therefore am I short
    Of knowing what I ought."
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  8. #8
    C++ Developer XSquared's Avatar
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    Code:
    #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...
    }
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