Category : argv

I have a program that basically looks like this: #include <iostream> #include <unistd.h> // getopt #include <vector> #include <string> #include <list> #include <sstream> using namespace std; //a parameter structure to store parameters provided through console typedef struct pairwise_param { double alpha; double beta; } param; //parse the parameter values void param_getopt(param *pm, int argc, char ..

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I executed the below program as: ./aout w. #include<iostream> using namespace std; int main(int argc, char** argv) { if (argv[1] == "w") { cout << "this was worked"; } else { cout << "this did not worked"; } } OUTPUT: this did not worked. I tried executing: if(&argv[1] == "w") if(argv[1]==’w’) Source: Windows Que..

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I am fairly new to C++ and am creating a project which sets your desktop wallpaper to whatever file you put in the directory. To get the current directory I am using argv[0]. This is, however, not returning the correct type to set the desktop wallpaper. Here is what I have. cout << argv[0] // ..

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I have a program that is similar to the following code int main(int argc, char** argv) { if (argv[2] == "x" || argv[2] == "y" || argv[2] == "z") { … } } When I run the program with $ program_name data z 0.1 0.4 It’s weird that the program never enters the if branch. ..

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I have code: int ParseCommandLine( int argc, const char* argv[]) { string inFilePath = ""; string outFilePath = ""; for( int i = 1; i < argc; ++i ) { if( string( argv[i] ) == "-i" || string( argv[i] ) == "–input") { // Check for "-i @args" form of reqest. if( argc <= 2 ..

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