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My argv[] in main always returns 0 when converted to integer


By : user2176420
Date : October 14 2020, 02:21 PM
I wish this help you argv is a pointer to char * which holds the arguments of the program.
*argv is equal to argv[0] which is the first argument, the program name. You actually want the argument argv[1] to be passed to atoi, but you should also check if the argument was passed:
code :
if(argc != 2)
{
    //print usage
    return 0;
}

int key = atoi(argv[1]);


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Checking if argv[i] is a valid integer, passing arguments in main


By : Alex Davis
Date : March 29 2020, 07:55 AM
I wish this help you I'm trying to make sure all arguments passed to main are valid integers, and if not, I'll print an error. For example, if I have an executable named total, I would enter total 1 2 3 4. I want to print an error if there's an invalid integer, so if I enter total 1 2 3zy it will print an error message. My code is as follows.
code :
bool legal_int(char *str)
{
 while(str != 0) // need to
 if( (isdigit(str)) )// do something here
 {
  return true;
 }
 else
 {
  return false;
 }
}

C++ int main(int argc, char* argv[]) arguments transform to integer values


By : WallyCode
Date : March 29 2020, 07:55 AM
With these it helps The problem is that there is no Console::WriteLine function taking a char*. You should convert it to a String object.
Regarding the warning message and the output, it's because since the compiler can't find an exact match for the WriteLine call, it will, if possible, pick another function. In this case it picks the one taking a Boolean argument, because pointers can implicitly be converted to bool which in turn can be converted to Boolean. The output is "true" because any non-null pointer is considered true in a pointer-to-bool conversion.

Get boost command line to parse array i supply rather than argv from main - emulate argv


By : LuLi
Date : March 29 2020, 07:55 AM
I hope this helps . You can't split into a char const*[].
Instead, split into a vector of std::string and transform to the required vector:
code :
#include <boost/program_options.hpp>
#include <boost/algorithm/string.hpp>
#include <iostream>

namespace po = boost::program_options;

int main()
{
    std::string input;

    while (std::cin) {
        std::getline(std::cin, input);

        std::vector<std::string> parsedInput;
        boost::split(parsedInput, input, boost::is_any_of(" "), boost::token_compress_on);

        std::vector<char const*> args { "command" };
        for (auto& arg : parsedInput)
            args.push_back(arg.c_str());

        po::options_description desc("allowed options");
        desc.add_options()
            ("help", "produce help message")
            ("compression", po::value<int>(), "set compression level")
        ;

        po::variables_map vm;
        po::store(po::parse_command_line(args.size(), args.data(), desc), vm);
        po::notify(vm);

        if (vm.count("help"))
            std::cout << desc << "\n";
    }
}
allowed options:
  --help                produce help message
  --compression arg     set compression level

tf.app.run(main=main, argv=[sys.argv[0]] + unparsed) AttributeError: module 'tensorflow' has no attribute 'app'


By : user3615010
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , Recording to the documentation of TensorFlow 2.0 the "app" attribuite has been removed: https://www.tensorflow.org/guide/effective_tf2
"Many APIs are either gone or moved in TF 2.0. Some of the major changes include removing tf.app, tf.flags, and tf.logging...". To resolve this, either uninstall tensorflow 2 and then install 1, or change your piece of code, I am not sure that going around this will help you alot, because Tensorflow is a very sensitive environment when you use it with Python, especially for the versions thing. so I think changing a little bit in the code would be better.

result of passing argv variable to main in this format main( int argc, char const * argv )


By : Jon Gilbert
Date : March 29 2020, 07:55 AM
I wish this help you It causes undefined behavior as the supplied type and the expected type does not match thereby this is a clear case of constraint violation.
Quoting C11, chapter chapter §5.1.2.2.1/p2, (emphasis mine)
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