Related
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
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
could you please help me understand this.
I have function that needs "char ***argv";
As far as I understand it's :
pointer to pointer to array of char pointers.
something like this: "char *arr[]" ?
char xx1 = '1';
char xx2 = '2';
char *argv[] = {&xx1,&xx2};
Then I call my function with gtk_init (&argc, &argv);
And get error:
main.cpp:43:31: error: cannot convert ‘char* (*)[2]’ to ‘char***’ for argument ‘2’ to ‘void gtk_init(int*, char***)’
Thank you for any help.
A char*** is a "pointer to pointer to pointer to char". No arrays involved at all. Your argv is an "array of 2 pointer to char". The name of your array, argv, will decay to a pointer to its first element in certain circumstances - this pointer will have type char**, or "pointer to pointer to char".
When you do &argv, you get a char* (*)[2], or a "pointer to array of 2 pointer to char". Which is not what you want. That's because you're taking the address of an array, not of a pointer.
Also, you're going to have a problem with the fact that you're pointers in argv are just pointing at single chars, not at null-terminated strings. gtk_init will almost certainly be expecting null-terminated strings.
So what you can you do? Try this:
char xx1[] = "1"; // Now these are null-terminated strings
char xx2[] = "2";
char* argv[] = {xx1, xx2};
char** argvp = argv; // Use the fact that the array will decay
gtk_init(&argc, &argv); // &argv will now be a char***
The reason for using arrays for the strings is because we need the chars to be non-const, but a char* str = "hello"; style declaration is deprecated - it must be const char*. However, by using arrays, the contents of the string literal are copied into our array, so we can freely make it non-const.
gtk_init really just expects you to pass the argc and argv parameters of your main function to it like so:
int main(int argc, char* argv[])
{
gtk_init(&argc, &argv);
// ...
}
You may now ask "But why is &argv now allowed? argv is the same type as in the question! Why don't we get a pointer to array again?" Actually, argv is not the same type, despite how much it looks like it is. When you define a function that takes an array argument, it actually gets transformed to a pointer. So the definition of main is equivalent to:
int main(int argc, char** argv);
So when we do &argv, we're totally fine, because it gives us a char***.
As #OmnipotentEntity says in the comments - you'd better have a good excuse for not passing the parameters of main to gtk_init.
What is the difference between char** argv and char* argv[]? in int main(int argc, char** argv) and int main(int argc, char* argv[])?
Are they the same? Especially that the first part does not have [].
They are entirely equivalent. char *argv[] must be read as array of pointers to char and an array argument is demoted to a pointer, so pointer to pointer to char, or char **.
This is the same in C.
They are indeed exactly the same.
The golden rule of arrays to remember is:
"The name of an array is a pointer to the first element of the array."
So if you declare the following:
char text[] = "A string of characters.";
Then the variable "text" is a pointer to the first char in the array of chars you've just declared. In other words, "text" is of type char *. When you access an element of an array using [index], what you're actually doing is adding an offset of index to the pointer to the first element of the array, and then dereferencing this new pointer. The following two lines will therefore initialize both variables to 't':
char thirdChar = text[3];
char thirdChar2 = *(text+3);
Using the square brackets is a convience provided by the language that makes the code much more readable. But the way this works is very important when you start thinking about more complex things, such as pointers to pointers. char** argv is the same as char* argv[] because in the second case "the name of the array is a pointer to the first element in the array".
From this you should also be able to see why it is that array indexes start from 0. The pointer to the first element is the array's variable name (golden rule again) plus an offset of... nothing!
I've had debates with a friend of mine as to which is better to use here. With the char* argv[] notation it may be clearer to the reader that this is in fact an "array of pointers to characters" as opposed to the char** argv notation which can be read as a "pointer to a pointer to a character". My opinion is that this latter notation doesn't convey as much information to the reader.
It's good to know that they're exactly the same, but for readablity I think that if the intention is an array of pointers then the char* argv[] notation conveys this much more clearly.
For the first part of the question:
char** argv: pointer to a pointer to a char
char* argv[]: pointer to an array
So the question is whether a pointer to a type C and an array C[] are the same things. They are not at all in general, BUT they are equivalent when used in signatures.
In other words, there is no difference in your example, but it is important to keep in mind the difference between pointer and array otherwise.
For all practical purposes, they're the same. This is due to C/C++'s handling of arrays passed as arguments, where an array decays to a pointer.
The bracket form is only useful in statement declarations like:
char *a[] = {"foo", "bar", "baz"};
printf("%d\n", sizeof a / sizeof *a);
// prints 3
Because it knows at compile time the size of the array. When you pass a bracket form as parameter to a function (main or some other one), the compiler has no idea what the size of the array would be at runtime, so it is exactly the same as char **a. I prefer char **argv since it's clearer that sizeof wouldn't work like it would on the statement declaration form.
There is a difference between TYPE * NAME and TYPE NAME[] in both C and C++. In C++ both types are not interchagneable. For example following function is illegal (you will get an error) in C++, but legal in C (you will get warning):
int some (int *a[3]) // a is array of dimension 3 of pointers to int
{
return sizeof a;
}
int main ()
{
int x[3][3];
std::cout << some(x)<< std::endl;
return 0;
}
To make it legal just change signature to int some (int (*a)[3]) (pointer to array of 3 ints) or int some (int a[][3]). The number in last square brackets must be equal to an argument's. Converting from array of arrays to an array of pointers is illegal. Converting from pointer to pointer to array of arrays is illegal too. But converting pointer to pointer to an array of pointers is legal!
So remember: Only nearest to dereference type signature doesn't matter, others do (in the context of pointers and arrays, sure).
Consider we have a as pointer to pointer to int:
int ** a;
&a -> a -> *a -> **a
(1) (2) (3) (4)
You cannot change this value, the type is int ***. May be taken by function as int **b[] or int ***b. The best is int *** const b.
The type is int **. May be taken by function as int *b[] or int ** b. Brackets of the array declaratin may be leaved empty or contain any number.
The type is int *. May be taken by function as int b[] or int * b or even void * b
Should be taken as int parameter. I don't want to fall into details, like implicit constructor call.
Answering your question: the real type of argumets in main function is char ** argv, so it may be easily represented as char *argv[] (but not as char (*argv)[]). Also argv name of main function may be safely changed.
You may check it easily: std::cout << typeid(argv).name(); (PPc = pointer to p. to char)
By the way: there is a cool feature, passing arrays as references:
void somef(int (&arr)[3])
{
printf("%i", (sizeof arr)/(sizeof(int))); // will print 3!
}
Moreover pointer to anything may be implicitly accepted (converted) by function as void pointer. But only single pointer (not pointer to pointer etc.).
Further reading:
Bjarne Stroustrup, C++, chapter 7.4
C pointers FAQ
This is a simple example I came up with, which have two functions (Main_1, Main_2) take the same arguments as the main function.
I hope this clear things up..
#include <iostream>
void Main_1(int argc, char **argv)
{
for (int i = 0; i < argc; i++)
{
std::cout << *(argv + i) << std::endl;
}
}
void Main_2(int argc, char *argv[])
{
for (int i = 0; i < argc; i++)
{
std::cout << *(argv + i) << std::endl;
}
}
int main()
{
// character arrays with null terminators (0 or '\o')
char arg1[] = {'h', 'e', 'l', 'l', 'o', 0};
char arg2[] = {'h', 'o', 'w', 0};
char arg3[] = {'a', 'r', 'e', '\0'};
char arg4[] = {'y', 'o', 'u', '\n', '\0'};
// arguments count
int argc = 4;
// array of char pointers (point to each character array (arg1, arg2, arg3 and arg4)
char *argPtrs[] = {arg1, arg2, arg3, arg4};
// pointer to char pointer array (argPtrs)
char **argv = argPtrs;
Main_1(argc, argv);
Main_2(argc, argv);
// or
Main_1(argc, argPtrs);
Main_2(argc, argPtrs);
return 0;
}
Output :
hello
how
are
you
hello
how
are
you
hello
how
are
you
hello
how
are
you
Both are same for your usage except for the following subtle differences:
Sizeof will give different results for both
Also second one may not be reassigned to new memory area since it's an array
With second one you can use only those indexes which are
valid. It's unspecified by C/C++ if you try to use an array index beyond
array length. However with char** you can use any index from 0 to ...
Second form can only be used as formal parameters to a function. While first can even be used to declare variables within a stack.
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