What Does Input Arguments is _

Question

I wanna know why is result 1 for no arguments.

int main(int argcount, char *arglist[]) {
    int i;
    printf("Number of arguments %d\n",argcount);
    printf("Arguments list:\n");
    for (int i=0;i<argcount;i++)
        printf("%s\n",arglist[i]);
    return EXIT_SUCCESS;
}

Answer 1

From the standard (C11), noting especially the bold bit:

If they are declared, the parameters to the main function shall obey the following constraints:

• The value of argc shall be nonnegative.
• argv[argc] shall be a null pointer.
• If the value of argc is greater than zero, the array members argv[0] through argv[argc-1] inclusive shall contain pointers to strings, which are given implementation-defined values by the host environment prior to program startup. The intent is to supply to the program information determined prior to program startup from elsewhere in the hosted environment. If the host environment is not capable of supplying strings with letters in both uppercase and lowercase, the implementation shall ensure that the strings are received in lowercase.
• If the value of argc is greater than zero, the string pointed to by argv[0] represents the program name; argv[0][0] shall be the null character if the program name is not available from the host environment. If the value of argc is greater than one, the strings pointed to by argv[1] through argv[argc-1] represent the program parameters.
• The parameters argc and argv and the strings pointed to by the argv array shall be modifiable by the program, and retain their last-stored values between program startup and program termination.

In other words, argc includes the argument representing the program name - the actual parameters to the program start at argv[1]. That becomes evident from the program output as per the following transcript, where the first argument is the program name:

pax> cat testprog.c
#include <stdio.h>
int main(int argc, char *argv[]) {
    printf("Argument count: %d\n", argc);
    printf("Arguments:\n");
    for (int i = 0; i < argc; i++)
        printf("   %s\n", argv[i]);
    return 0;
}

pax> gcc --std=c11 -o testprog testprog.c && ./testprog 1 2 3
Argument count: 4
Arguments:
   ./testprog
   1
   2
   3

pax> ./testprog
Argument count: 1
Arguments:
   ./testprog