C++ Tutorial/Development/command line arguments — различия между версиями
Admin (обсуждение | вклад) м (1 версия: Импорт контента...) |
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(нет различий)
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Текущая версия на 10:28, 25 мая 2010
Display command-line arguments
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
for(int i = 0; i < argc; i++) {
cout << argv[i] << "\n";
}
return 0;
}main
Sum the two numeric command line arguments
#include <iostream>
#include <cstdlib>
using namespace std;
int main(int argc, char *argv[])
{
double a, b;
if(argc!=3) {
cout << "Usage: main num num\n";
return 1;
}
a = atof(argv[1]); // convert first command-line arg
b = atof(argv[2]); // convert second comnand-line arg
cout << a + b;
return 0;
}Usage: main num num
Using command-line arguments for copying files
#include <iostream>
using std::cout;
using std::endl;
using std::ios;
#include <fstream>
using std::ifstream;
using std::ofstream;
int main( int argc, char *argv[] )
{
// check number of command-line arguments
if ( argc != 3 )
cout << "Usage: copyFile infile_name outfile_name" << endl;
else
{
ifstream inFile( argv[ 1 ], ios::in );
if ( !inFile )
{
cout << argv[ 1 ] << " could not be opened" << endl;
return -1;
}
ofstream outFile( argv[ 2 ], ios::out );
if ( !outFile )
{
cout << argv[ 2 ] << " could not be opened" << endl;
inFile.close();
return -2;
}
char c = inFile.get();
while ( inFile )
{
outFile.put( c );
c = inFile.get();
}
}
return 0;
}Usage: copyFile infile_name outfile_name
