So now I have a
int main (int argc, char *argv[]){}
how to make it string based? will int main (int argc, std::string *argv[]) be enough?
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6 Answers
You can't change main's signature, so this is your best bet:
#include <string>
#include <vector>
int main(int argc, char* argv[])
{
std::vector<std::string> params(argv, argv+argc);
// ...
return 0;
}
If you want to create a string out of the input parameters passed, you can also add character pointers to create a string yourself
#include <iostream>
#include <string>
using namespace std;
int main(int argc, char* argv[])
{
string passedValue;
for(int i = 1; i < argc; i++)
passedValue += argv[i];
// ...
return 0;
}
1 Comment
Aaron D. Marasco
The "
+=" would mash them all together. passedValue.append(argv[i]).append(" ") would look nicer.You can't do it that way, as the main function is declared explicitly as it is as an entry point. Note that the CRT knows nothing about STL so would barf anyway. Try:
#include <string>
#include <vector>
int main(int argc, char* argv[])
{
std::vector<std::string> args;
for(int i(0); i < argc; ++i)
args.push_back(argv[i]);
// ...
return(0);
}; // eo main
3 Comments
Chubsdad
vow. like that i(0) syntax +1
luke
Since you know the size of the resulting vector, you could use vector's size constructor to preallocate, avoiding possible multiple reallocations with .push_back(). You could also use vector's iterator constructor to avoid having to write the loop yourself.
Moo-Juice
@luke, yup I saw your response and +1'd it as it is far cleaner than mine.
That would be non-standard because the Standard in 3.6.1 says
An implementation shall not predefine the main function. This function shall not be overloaded. It shall have a return type of type int, but otherwise its type is implementation-defined. All implementations shall allow both of the following definitions of main:
int main() { /* ... */ }and
int main(int argc, char* argv[]) { /* ... */ }
1 Comment
Prasoon Saurav
@Chubsdad : I use Foxit
;-)No. That is not allowed. If present, it is mandated to be char *argv[].
BTW, in C++ main should always be declared to return 'int'
Comments
main receives char*. you will have to put the argv array into std::strings yourself.
1 Comment
Chubsdad
main does not pass anything. instead, main is passed char* by the CRT