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FrEqDe
Asked: 2020-03-14 18:31:37 +0800 CST 2020-03-14 18:31:37 +0800 CST

What is the difference between a main with arguments and another without arguments in c++?

  • 772

There is a main that is like this.

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

}

And another one that is just like this:

int main(){

  }

So far I have used both and both have worked for me, so I have not identified what the difference is between using one or the other, or if they have different purposes.

c++

3 Answers

  1. Best Answer

    In the firstmain

    • argc: stores the amount (number) of parameters, including the name of the executable.

    • *argv[]: An array containing all received parameters 1 including the name of the executable.

    if you use terminal many times you do things like this:

    miPrograma argumento1 argumento2

    that's when through the first mainbut NOT the second, you can get those parameters and use them within your program to generate one behavior or another, check what parameter was passed, or the number of parameters that have been passed when executing your program etc.

    you can compile this example:

    int main(int argc, char *argv[])
{
    printf("numero de argumentos: %i\n", argc);

    printf("Argumento 0: %s\n", argv[0]);
    printf("Argumento 1: %s\n", argv[1]);
    printf("Argumento 2: %s\n", argv[2]);
    printf("Argumento 3: %s\n", argv[3]);
    return 0;
}

    and run it like this for example:

    a.out stack overflow es

    Shell:

    ./a.out stack overflow es
numero de argumentos: 4
Argumento 0: ./a.out
Argumento 1: stack
Argumento 2: overflow
Argumento 3: es

    (if your program is called a.out)

    the following code uses this syntax char **argvinstead of char *argv[]but is "basically" the same. One is more than c and the other is c++, some call it a two-dimensional array of characters and others an array of pointers to characters .

    #include <stdio.h>

int main(int argc, char **argv)
{
    printf("Argumento 0: %s\n", argv[0]);
    printf("Argumento 1: %s\n", argv[1]);
    printf("Argumento 2: %s\n", argv[2]);
    printf("Argumento 3: %s\n", argv[3]);
    return 0;
}

    1 *argv[]: array of pointers to characters.

    • 5

  2. What is the difference between a main with arguments and another without arguments in c++?

    The question itself has the answer: The arguments. Both versions will be considered the entry point of the program.

    According to the C++ standard (my translation):

    3.6.1 main function

    1. The program must contain a global function called main, which will be the starting point of the program. [...]

    2. The implementation must not predefine the function main. This function should not be overloaded. Must be declared with a return value int[...] The implementation must allow both signatures

      2.1. - a function ()returning inty

      2.2. - a function (int,pointer to pointer to char)returningint

      like the type of main.

    Thus, both signatures are correct and valid for mainand the standard does not indicate that either version of mainshould act differently.

    • 2

  3. Both forms are valid and are the entry point for your application.

    The first form int main(int argc, char** argv)receives the command line arguments (when the application is launched), in the form of: array of null-terminated strings (argv) and the number of strings in that array (argc).

    The second way int main()is simpler, preferred when the application doesn't need to read the command line arguments (ie, doesn't use them).

    Although you haven't mentioned it, it mainshould return 0 if execution was successful and non-0 if there was a failure. If you omit the return, c++ assumes that you returned 0 (or success).

    • 1