how to point to char *argv[] in main? - c++

In c++ I've got a main function with
int argc, char * argv[]
I need to access the data in argv[] (i.e the arguments) in another function.
I am going to declare a global variable, a pointer to the char **argv.
How do I do this?


In C++, usually* the best way to handle argv is this:
int main(int argc, char **argv)
{
std::vector<std::string> args(argv, argv + argc);
}
Now you have args, a correctly constructed std::vector holding each element of argv as a std::string. No ugly C-style const char* in sight.
Then you can just pass this vector or some of its elements as you need.
* It carries the memory & time overhead of dynamically allocating one copy of each argv string. But for the vast majority of programs, command-line handling is not performance-critical and the increased maintainability and robustness is well worth it.


Global variables should be avoided and you should prefer passing arguments. Anyway, you can just use:
char **global_argv = NULL;
int main(int argc, char * argv[]){
...
global_argv = argv;
...
}
A better way would be:
void my_fun1(int argc, char **argv); // Passing all the program arguments
void my_fun2(char *arg); // Passing just the one you need
int main(int argc, char * argv[]){
...
my_fun1(argc, argv);
my_fun2(argv[3]); // Supposing you need the 3rd parameter (fourth on argv)
...
}

Related

What is the difference between `char **argv` and `char** argv`? [duplicate]

This question already has answers here:
Declaring pointers; asterisk on the left or right of the space between the type and name? [duplicate]
(6 answers)
Closed 1 year ago.
I have the two following C++ main function headers:
int main(int argc, char **argv)
int main(int argc, char** argv)
I know that the asterisk * signifies a pointer, but I do not understand the difference between these two. The one I usually see is char **argv, which, If I'm not mistaken, declares argv to be a pointer to a pointer to a char. If so, then what is char** argv?

If so, then what is char** argv?
If (also) declares argv to be a pointer to a pointer to a char.
What is the difference between char **argv and char** argv?
The difference between them is the placement of space. There is no other difference. These are also the same:
char**argv
char * * argv
char
*
*
argv

C++ editing the elements of char* argv[]

All of you know this function:
int main(int argc, char* argv[])
{
}
I want to write a command line interface in Linux for my program, which is usually done by getopt_long()
My program would be executed from command line like this:
pop3 get --limit 25 --recent
Hence, the argv[] would include pop3 as its program name, and the rest are treated as options.
I want to delete pop3 from my string and set the first token after it as the first element of the array. Is there a way to do that other than looping through?

Increment the argv pointer, and decrement the argc.
Example:
int main(int argc, char *argv[])
{
argc--;
argv++;
return 0;
}
This works, because when you increment argv, you still have the previous data in memory, it's just that the base address of the argv has increased. And you decrement argc, because you now have one less argument.

What is a char** argv[] in C++?

I am clear about the difference between char* argv[] and char** argv (as explained, for instance, in this question).
But what kind of type is char** argv[] ? A pointer to a pointer to a pointer of chars? And what would that be, when used to pass arguments to the program?
For the record, the type occurs in the following declaration of a C++ library I am trying to interface to Python:
/** Initialize the Stage library. Stage will parse the argument
array looking for parameters in the conventional way. */
void Init( int* argc, char** argv[] );

If main definition is main(int argc, char * argv[]) and you want to send argv via pointer, you will do Init( &argc, &argv ). And the corresponding parameter can be char ** argv[].

I assume that this Init function will make changes to the argc and argv that main has received, so it needs pointers to them.
If the library uses some program arguments, the Init function can handle them, and then remove those arguments from the argument list before it is processed by the rest of the program.

char** argv[] is an array of char**, IOW an array of pointers to pointers to chars, or more likely an array of pointers to null-terminated character strings, eg:
char *str1 = "...";
char *str2 = "....";
char** argv[2];
argv[0] = &str1;
argv[1] = &str2;
However, if argv is coming from main() itself, then char** is a typo, it must be char* instead:
char* argv[]
But if argv is coming from some library instead, it very well could be wrapping the original argv array from main() inside its own array, in which case char**[] might make sense so it can point at the original strings and not have to copy them.

C++ int main (int argc, char *argv[]) - is argv a c style array?

the following does not compile:
int main (int argc, char *argv[]) {
char arr[2][2];
char **ptr = arr;
return 0;
}
Yet the following does:
int main (int argc, char *argv[]) {
char **ptr = argv;
return 0;
}
Thus I would conclude that argv is not a c style array which is stored as a single block of memory but rather is some sort of pointer array. Can somebody confirm that?
Thank you and Regards.

That's correct, argv is a pointer to the first element of an array of pointers, each of which in turn (except for the last one) is a pointer to the first character in a string of characters giving the respective command line argument.
Function parameters cannot be arrays.

What does int argc, char *argv[] mean?

In many C++ IDE's and compilers, when it generates the main function for you, it looks like this:
int main(int argc, char *argv[])
When I code C++ without an IDE, just with a command line compiler, I type:
int main()
without any parameters. What does this mean, and is it vital to my program?

argv and argc are how command line arguments are passed to main() in C and C++.
argc will be the number of strings pointed to by argv. This will (in practice) be 1 plus the number of arguments, as virtually all implementations will prepend the name of the program to the array.
The variables are named argc (argument count) and argv (argument vector) by convention, but they can be given any valid identifier: int main(int num_args, char** arg_strings) is equally valid.
They can also be omitted entirely, yielding int main(), if you do not intend to process command line arguments.
Try the following program:
#include <iostream>
int main(int argc, char** argv) {
std::cout << "Have " << argc << " arguments:" << std::endl;
for (int i = 0; i < argc; ++i) {
std::cout << argv[i] << std::endl;
}
}
Running it with ./test a1 b2 c3 will output
Have 4 arguments:
./test
a1
b2
c3

argc is the number of arguments being passed into your program from the command line and argv is the array of arguments.
You can loop through the arguments knowing the number of them like:
for(int i = 0; i < argc; i++)
{
// argv[i] is the argument at index i
}

Suppose you run your program thus (using sh syntax):
myprog arg1 arg2 'arg 3'
If you declared your main as int main(int argc, char *argv[]), then (in most environments), your main() will be called as if like:
p = { "myprog", "arg1", "arg2", "arg 3", NULL };
exit(main(4, p));
However, if you declared your main as int main(), it will be called something like
exit(main());
and you don't get the arguments passed.
Two additional things to note:
These are the only two standard-mandated signatures for main. If a particular platform accepts extra arguments or a different return type, then that's an extension and should not be relied upon in a portable program.
*argv[] and **argv are exactly equivalent, so you can write int main(int argc, char *argv[]) as int main(int argc, char **argv).

int main();
This is a simple declaration. It cannot take any command line arguments.
int main(int argc, char* argv[]);
This declaration is used when your program must take command-line arguments. When run like such:
myprogram arg1 arg2 arg3
argc, or Argument Count, will be set to 4 (four arguments), and argv, or Argument Vectors, will be populated with string pointers to "myprogram", "arg1", "arg2", and "arg3". The program invocation (myprogram) is included in the arguments!
Alternatively, you could use:
int main(int argc, char** argv);
This is also valid.
There is another parameter you can add:
int main (int argc, char *argv[], char *envp[])
The envp parameter also contains environment variables. Each entry follows this format:
VARIABLENAME=VariableValue
like this:
SHELL=/bin/bash
The environment variables list is null-terminated.
IMPORTANT: DO NOT use any argv or envp values directly in calls to system()! This is a huge security hole as malicious users could set environment variables to command-line commands and (potentially) cause massive damage. In general, just don't use system(). There is almost always a better solution implemented through C libraries.

The parameters to main represent the command line parameters provided to the program when it was started. The argc parameter represents the number of command line arguments, and char *argv[] is an array of strings (character pointers) representing the individual arguments provided on the command line.

The main function can have two parameters, argc and argv. argc is an integer (int) parameter, and it is the number of arguments passed to the program.
The program name is always the first argument, so there will be at least one argument to a program and the minimum value of argc will be one. But if a program has itself two arguments the value of argc will be three.
Parameter argv points to a string array and is called the argument vector. It is a one dimensional string array of function arguments.

Lets consider the declaration:
int main (int argc, char *argv[])
In the above declaration, the type of the second parameter named argv is actually a char**. That is, argv is a pointer to a pointer to a char. This is because a char* [] decays to a char** due to type decay. For example, the below given declarations are equivalent:
int main (int argc, char *argv[]); //first declaration
int main (int argc, char **argv); //RE-DECLARATION. Equivalent to the above declaration
In other words, argv is a pointer that points to the first element of an array with elements of type char*. Moreover, each elements argv[i] of the array(with elements of type char*) itself point to a character which is the start of a null terminated character string. That is, each element argv[i] points to the first element of an array with elements of type char(and not const char). A diagram is given for illustration purposes:
As already said in other answers, this form of declaration of main is used when we want to make use of the command line argument(s).

The first parameter is the number of arguments provided and the second parameter is a list of strings representing those arguments.

Both of
int main(int argc, char *argv[]);
int main();
are legal definitions of the entry point for a C or C++ program. Stroustrup: C++ Style and Technique FAQ details some of the variations that are possible or legal for your main function.

In case you learn something from this
#include<iostream>
using namespace std;
int main(int argc, char** argv) {
cout << "This program has " << argc << " arguments:" << endl;
for (int i = 0; i < argc; ++i) {
cout << argv[i] << endl;
}
return 0;
}
This program has 3 arguments. Then the output will be like this.
C:\Users\user\Desktop\hello.exe
hello
people

When using int and char**, the first argument will be the number of commands in by which the programs is called and second one is all those commands

Just to add because someone says there is a third parameter (*envp[]), it's true, there is, but is not POSIX safe, if you want your program to use environment variables you should use extern char environ ;D