*++argv[0] construct in C language

Question:

How does the *++argv[0] construct work in the inner loop in this example of using pointers?

#include <string.h>
#include <stdio.h>

int main(int argc, char *argv[])
{
        int c, except = 0, number = 0;

        while(--argc > 0 && (*++argv)[0] == '-')
        {
                printf("argv = %s\n", *argv);
                while(c = *++argv[0])
                        switch(c){
                                case 'x':
                                        except = 1;
                                        break;
                                case 'n':
                                        number = 1;
                                        break;
                                default:
                                        printf("Illegal option: %c\n", c);
                                        argc = 0;
                                        break;
                        }

        }

        printf("[DEBUG]: argc = %d\n", argc);

        if(argc != 1)
                printf("Usage: -x -n\n");
        else
                printf("x = %d, n = %d\n", except, number);
        return 0;
}

I will analyze a similar construction (*++argv)[0] in the outer loop. This takes the null character of the string that contains one of the arguments. Brackets [] have higher precedence than increment and reference, so *++argv is enclosed in brackets. The zero element of the array argv is a pointer to the name of the program, so we increment to the next line. After that, we dereference the pointer and get the string. Using brackets [0] gives the null character of this string.

The construction *++argv[0] is completely incomprehensible to me. There should be a pass along the argument string. This is necessary in order to support not only keys like -x -n , but also keys like -nx . What is [0] here for? How does an entire expression work?

Answer:

Well, here it's just a character-by-character pass – as you wrote yourself, square brackets have a higher precedence, so the argv[0] pointer, which initially points to the beginning of the argument string, is simply incremented, and dereferencing gives the next character…

Those. if the program is called such that argv[1] == -abcxn , then after the first line (++argv)[0] the pointer argv[0] points to the line -abcxn . The first dereference (in the first while ) gives '-' , then, in the second while , it just loops over all the characters in the string – a , b , c , x , n .

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