How to cast program argument *char as int correctly

ccastingcharpointers

I'm on a Mac OS X 10.6.5 using XCode 3.2.1 64-bit to build a C command line tool with a build configuration of 10.6 | Debug | x86_64. I want to pass a positive even integer as an argument, so I'm casting index 1 of argv as an int. This seems to work, except it seems that my program is getting the ascii value instead of reading the whole char array and converting to an int value. When I enter 'progname 10', it tells me that I've entered 49, which is what I also get when I enter 'progname 1'. How can I get C to read the entire char array as an int value? Google only showed me (int)*charPointer, but clearly that doesn't work.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int main(int argc, char **argv) {
    char *programName = getProgramName(argv[0]); // gets the name of itself as the executable
    if (argc < 2) {
        showUsage(programName);
        return 0;
    }
    int theNumber = (int)*argv[1];
    printf("Entered [%d]", theNumber);            // entering 10 or 1 outputs [49], entering 9 outputs [57]
    if (theNumber % 2 != 0 || theNumber < 1) {
        showUsage(programName);
        return 0;
    }

    ...

}

Best Answer

The numer in the argv array is in a string representation. You need to convert it to an integer.

sscanf

int num;
sscanf (argv[1],"%d",&num);

or atoi() (if available)

Related Solutions

C# – How to cast int to enum

From an int:

YourEnum foo = (YourEnum)yourInt;

From a string:

YourEnum foo = (YourEnum) Enum.Parse(typeof(YourEnum), yourString);

// The foo.ToString().Contains(",") check is necessary for enumerations marked with an [Flags] attribute
if (!Enum.IsDefined(typeof(YourEnum), foo) && !foo.ToString().Contains(","))
{
    throw new InvalidOperationException($"{yourString} is not an underlying value of the YourEnum enumeration.")
}

Update:

From number you can also

YourEnum foo = (YourEnum)Enum.ToObject(typeof(YourEnum) , yourInt);

C++ – How to convert a std::string to const char* or char*

If you just want to pass a std::string to a function that needs const char* you can use

std::string str;
const char * c = str.c_str();

If you want to get a writable copy, like char *, you can do that with this:

std::string str;
char * writable = new char[str.size() + 1];
std::copy(str.begin(), str.end(), writable);
writable[str.size()] = '\0'; // don't forget the terminating 0

// don't forget to free the string after finished using it
delete[] writable;

Edit: Notice that the above is not exception safe. If anything between the new call and the delete call throws, you will leak memory, as nothing will call delete for you automatically. There are two immediate ways to solve this.

boost::scoped_array

boost::scoped_array will delete the memory for you upon going out of scope:

std::string str;
boost::scoped_array<char> writable(new char[str.size() + 1]);
std::copy(str.begin(), str.end(), writable.get());
writable[str.size()] = '\0'; // don't forget the terminating 0

// get the char* using writable.get()

// memory is automatically freed if the smart pointer goes 
// out of scope

std::vector

This is the standard way (does not require any external library). You use std::vector, which completely manages the memory for you.

std::string str;
std::vector<char> writable(str.begin(), str.end());
writable.push_back('\0');

// get the char* using &writable[0] or &*writable.begin()