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Pass arguments into C program from command line.
mypro parameter
When run like above,how to get the parameter in mypro's main() :
#include <iostream>
int main()
{
char* str = "default_parameter";
if(parameter_exists())str = parameter;
...
}
How to implement the pseudo code above?
Just need to add (int argc, char *argv[]) to your main function. argc holds the number of arguments, and argv the arguments themselves.
int main(int argc, char *argv[])
{
std::string str = "default";
if (argc > 1) { str = argv[1]; }
}
Note that the command is also included as an argument (e.g. the executable). Therefore the first argument is actually argv[1].
When expecting command-line arguments, main() accepts two arguments: argc for the number of arguments and argv for the actual argument values. Note that argv[0] will always be the name of the program.
Consider a program invoked as follows: ./prog hello world:
argc = 3
argv[0] = ./prog
argv[1] = hello
argv[2] = world
Below is a small program that mirrors the pseudocode:
#include <iostream>
#include <string>
int main(int argc, char **argv) {
std::string arg = "default";
if (argc >= 2) {
default = argv[1]
}
return 0;
}
Your function should use the second of these two allowed C++ main function signatures:
int main(void)
int main(int argc, char *argv[])
In this, argc is the argument count and argv is an argument vector.
For more details, see this Wikipedia article.
Well you need a main function that accepts arguments, like so:
int main(int argc, char *argv[])
{
}
You would then check the number of arguments passed in using argc. Usually there is always one, the name of the executable but I think this is OS dependent and could thus vary.
You need to change the main to be:
int main(int argc, char** argv){
argc is the number of paramaters, argv is an array containing all of the paramaters (with index 0 being the name of the program). Your first parameter will be located at index 1 of argv.
#include <iostream>
using namespace std;
int main(int argc, char* argv[]) {
for(int i = 1; i < argc; i++)
cout << atoi(argv[i]) << endl;
return 0;
}
Here argc gives the count of the arguments passed
and argv[i] gives ith command line parameters.
where argv[0]='name of the program itself'
main has two parameters, int argc and char** argv which you can use to access to the command line parameters.
argc is the number of parameters including the name of the executable, argv points to the parameter list itself (so that argv[0] is "mypro" in your example).
Just declare main like
int main (int argc, char** argv){..}
You need to use this declaration of main:
int main(int argc, _TCHAR* argv[])
{
if (argc > 1)
{
str = argv[1];
}
}
argv[0] is the name of the executable, so the parameters start at argv[1]
On Windows, use the following signature instead to get Unicode arguments:
int wmain(int argc, wchar_t** argv)
Related
Is there a way to keep the command line argument as standard and yet pass the argv internally inside the main?
Like, change this:
int main(int argc, char **argv) {
App app(argc,argv);
return app.exec();
}
to something like this:
int main(int argc, char **argv) {
vector<string> argv ="arguments";
int argc = X;
App app(argc,argv);
return app.exec();
}
When I have:
./a.out -a A -b B -c C
I know this is weird. I just want to find a workaround not to change the source code of a huge project and just run my command line arguments every time with only ./a.out.
Put each of your arguments in a char array, and then put pointers to those arrays into an array of pointers.
char arg1[] = "./a.out";
...
char argN[] = "whatever";
char* argv[] = { arg1, ..., argN}
App app(N, argv);
You may be looking for
const char *myArgv[]={"-a","A","-b","B"};
int myArgc=4;
App app(myArgc,myArgv);
return app.exec();
std::vector<std::string> args(argv, argv+argc);
You may rename arguments passed to app as you wish to avoid a name conflict, for ex.
int main(int argc, char **argv) {
vector<string> app_argv = /* contents */;
int app_argc = app_argv.size() ;
App app(app_argc, app_argv);
return app.exec();
}
How do I read command-line parameters in C? For example, in
./test --help
or
./test --build
how do I access "--build" or "--help"?
Your parameters are in argv:
int main(int argc, char **argv)
if you printf the content of argv (argv[0],argv[1] etc) youll get the idea.
try:
int main (int argc, char **argv)
{
for(int i = 0;i< argc;i++)
printf("%s\r\n",argv[i]);
}
You can use the argc and argv arguments to the main function and do different things based on them:
#include <string.h>
void dohelp(void) { /* whatever; */ }
void dobuild(void) { /* whatever; */ }
int main(int argc, char **argv) {
if (argc == 2) {
if (!strcmp(argv[1], "--help")) dohelp();
if (!strcmp(argv[1], "--build")) dobuild();
}
return 0;
}
argc contains the number of parameters passed by the shell to your program, including the program name. So myapp --help gets an argc of 2.
argv are the arguments themselves. The last argv (argv[argc]) is the NULL pointer.
Edit: the parameters don't need to be named argc and argv, but naming else something else is very, very bad!
int main(int foo, char **bar) /* RGAGGGGHH */
int main(int n, char **options) /* RGAGGGGHH */
The very basic is to use the arguments (int argc, char *argv[]) and you can parse those directly.
One more advanced method is to use getopt... http://www.gnu.org/s/libc/manual/html_node/Getopt.html
Command line arguments are arguments passed to your program with its name. For example, the UNIX program cp (copies two files) has the following command line arguments:
cp SOURCE DEST
You can access the command line arguments with argc and argv:
int main(int argc, char *argv[])
{
return 0;
}
argc is the number of arguments, including the program name, and argv is the array of strings containing the arguments. argv[0] is the program name, and argv[argc] is guaranteed to be a NULL pointer.
So the cp program can be implemented as such:
int main(int argc, char *argv[])
{
char *src = argv[1];
char *dest = argv[2];
cpy(dest, src);
}
They do not have to be named argc and argv; they can have any name you want, though traditionally they are called that.
There are several ways to do it [as usual].
Command line arguments are read from argv (passed to main along with argc).
You can parse those yourself and have a bit switch setting flags each time a new option is found in argv. Or you can use a library to parser command line arguments. I suggest libc getopt (google it).
I have a program that requires a command line argument to run properly, but it runs even when no argument is provided. How do I ensure that an argument is provided before it runs?
int main(int argc, const char ** argv) {
std::ifstream b(argv[1]);
Word c;
c.fillWords(c.getWordsAdress(), &b);
c.printWord(c.getWordsAdress());
}
Check the argument count like this:
int main( int argc, const char* argv[] )
{
if (argc < 2)
return 1;
// your code here
}
You can just check the argument count and if it is less than 2, that means that no arguments were provided. The argument count will always have at least 1, which contains the name of the program.
int main(int argc, char** argv)
{
if(argc < 2) {
cerr << "usage: " << argv[0] << " -argument";
return EXIT_FAILURE;
}
return EXIT_SUCCESS;
}
#include <stdio.h>
int main(char sendbuf[100])
{
printf (sendbuf);
return 0;
}
Somehow this very basic program crashes when I try to use it, it's meant to print up whatever is typed as a parameter. If I remove the line "printf (sendbuf);" the crash goes away.
The first argument to main is the number of parameters. The second argument is an array of strings. The first element (index 0) of the second argument is the name of your program:
#include <stdio.h>
int main(int c, char **argv)
{
printf ("%s\n", c > 1 ? argv[1] : "No Argument");
return 0;
}
Your first parameter must be an integer, not a char array. Here is the right program:
#include <stdio.h>
int main(int argc, char* argv[])
{
if (argc > 1) {
printf( argv[1] );
}
else {
printf( "No arguments provided" );
}
return 0;
}
argv[0] is your program name, so argv[1] is the first parameter provided on teh command line.
C supports two forms of main function:
int main() { /* ... */ }
and
int main(int argc, char* argv[]) { /* ... */ }
To take parameter from main, you need to change your code to:
#include <stdio.h>
int main(int argc, char* argv[])
{
if (argc > 1){
printf ("%s\n", argv[0]);
}
return 0;
}
Or use stream:
#include <iostream>
int main(int argc, char* argv[])
{
if (argc > 1){
std::cout << argv[0]) << std::endl;
}
return 0;
}
argv[0] is application name, input parameters starts from argv[1] if any.
An implementation must support the following two definitions of main:
int main() { }
int main(int argc, char* argv[]) { }
It is implementation-defined whether they support any other definitions. I don't know of any implementation that allows int main(char*) though (which is what yours is equivalent to).
This will print everything you type on the command line after the program name, even with spaces. It will not crash if you type nothing after the program name.
#include <stdio.h>
int main(int argc, char **argv)
{
for(int i=1; i<=argc; ++i) {
printf("%s\n", argv[i]);
}
}
Sorry for such a noobie question, but how can I get a program to read the data i input with my program, like how cmd does it with the options
shutdown.exe -f
how do i read the example -f into my program?
This should print out each of the whitespace delimited parameters which were passed to your program.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
for(int i = 0; i < argc; i++)
{
printf("%s\n", argv[i]);
}
return 0;
}
If you're using plain old C++, your main function will need to look something like this:
int main(int argc, char *argv[])
where argc is the number of whitespace separated items, and argv is an array of pointers to each one
int main(int argc, char *argv[])
{
}
arguments are passed by argv.