int main（int argc，char * argv []）

Question

If I have this:

int main(int argc, char *argv[])

In the body, you can sometimes find programs using argv[1] .

When do we use argv[1] over argv[0] ? Is it only when we just want to read the second argument in the command line?

Answer 1

By convention , argv[0] is the current program's name (or path), and argv[1] through argv[argc - 1] are the command-line arguments that the user provides.

However, this doesn't have to be true -- programs can OS-specific functions to bypass this requirement, and this happens often enough that you should be aware of it. (I'm not sure if there's much you can do even if you're aware of it, though...)

Example:

gcc -O3 -o temp.o "My file.c"

would (should) produce the following arguments:

argc: 5
argv: ["gcc", "-O3", "-o", "temp.o", "My file.c"]

So saying argv[0] would refer to gcc , not to -O3 .

Answer 2

argv is an array of pointers, and each pointer in this array stores one argument from command line. So argv[0] is the first argument (that is the executable/program itself), argv[1] is the second argument, and so on!

The total number of arguments is determined by argc .

Answer 3

Let's suppose your C++ executable file is:

/home/user/program (or C:\\program.exe in Windows)

if you execute:

./home/user/program 1 2 (or C:\\program.exe 1 2 in Windows)

argv[0] = /home/user/program ( C:\\program.exe )
argv[1] = 1
argv[2] = 2

That is because:

• argv[0] is the path of the executable file
• argv[1] is the 1st argument

Edit:

Now I see that argv[0] isn't necessarily the path of the executable file.
Read the following SO question: Is args[0] guaranteed to be the path of execution?

Answer 4

argv [0]是程序的执行路径，argv [1]是程序的第一个参数

Answer 5

Yes, that's mostly it, argv[1] is the second command line parameter. The first command line parameter is the name of the program itself.

Alternatively, to avoid the semantic mess that this answer originally had, and the comments from others, it might make sense to call argv[0] the zeroth parameter, so that argv[1] would now be the "first" of the user supplied values.

In any event, this comes from the exec() family of functions, eg execl which has usage:

 int execl(const char *path, const char *arg0, ... /*, (char *)0 */);

In the (Unix) shell when you type in a command, if necessary the shell first resolves the command name (using $PATH ) to find the real absolute path. The (absolute or relative) path is supplied for path , and the command as originally typed-in is supplied as arg0 , eventually becoming argv[0] in your program.

The remaining command line parameters then end up as argv[1] , etc.

Answer 6

简短回答是，数组包含传递给程序的所有选项。

Answer 7

as argv[0] is filepath of the program itself. Extra command line parameters are in further indexes, argv[1],argv[2].. You can read more here : http://www.site.uottawa.ca/~lucia/courses/2131-05/labs/Lab3/CommandLineArguments.html