Argv and argc in C

Question

I have problem with argc.

Command line I sent: prog 333 jos nije kraj akademske godine

void main(int argc, char *argv[]){
    char pom[40];
    char *ptr;
    strcpy(pom, argv[5]+3);
    printf("%s;", pom);
    ptr=strchr(argv[6], 'd');
    printf("%s;", ptr);
    printf("%d.", argc);

    return 0;
}

I get j;demske;8.

Why is it 8 and not 7. I should get number of parameters sent.

is `prog` the first argument or the name of your program? – Gerhardh Jun 22 '20 at 08:52 — Gerhardh, Jun 22 '20 at 08:52
Did you try to print all the arguments you get? – Gerhardh Jun 22 '20 at 08:53 — Gerhardh, Jun 22 '20 at 08:53

score 1 · Answer 1 · answered Jun 22 '20 at 08:55

1

"akademske" is argv[6] so you have arguments argv[0] to argv[7]. So these are 8 arguments. argv[0] is the name of your program.

answered Jun 22 '20 at 08:55

Jens Gustedt

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score 1 · Answer 2 · answered Jun 22 '20 at 08:57

Taking into account the program output it seems that these sequence of wards

 prog 333 jos nije kraj akademske godine

is command line arguments. So including the program name itself that corresponds to the argv[0] argc is equal to 8.

From the C Standard (5.1.2.2.1 Program startup)

If the value of argc is greater than zero, the string pointed to by argv[0] represents the program name;

argv[5] is the string kraj. So these statements

strcpy(pom, argv[5]+3);
printf("%s;", pom);

output 'j'.

And these statements

ptr=strchr(argv[6], 'd');
printf("%s;", ptr);

output demske.

Argv and argc in C

2 Answers2