C Programming Quiz #2

[neothelord]

Guys,

The second quiz in C programming (10 multiple choice questions with feed back) is ready.

http://mulogic.phpnet.us/drupal/?q=C+programming-Quiz+2

Rate your C skills.

[nixen]

sorry my bad english but i can't belive that i'm studying c++ language for over 2 year at school and i nerver see code like this :
void main()
{
char *s[]={ "miller","miller","filler","filler"};
char **p;
p=s;
printf("%s ",++*p);
printf("%s ",*p++);
printf("%s ",++*p);
}

can you explain me how it works? thanks

[neothelord]

here s is an array of 4 character pointers. so when you refer an array by base name, it decays to a pointer to the first element . hence basename s is char **.

so ptr is a pointer to the first element of the array. when you dereference it with a *, you get the value stored in the first element of the array, which is the pointer to the first string "miller". The prefix ++, increments this pointer by 1. since the data type is char *, the pointer is incremented to point the next byte(second char of the string "miller"). and the first printf prints "iller"

In the second printf postfix ++ has more precedence than *. so it increments ptr to point to second element of s. but remember, this side effect is applicable only after a sequence point. so the *ptr, still yields the value at the first element of s.

when control comes to the third ptr, ptr is pointing to the second elemtn of s. here we do a normal dereference to get the address of the second string "miller", and increment it by a char. so this also prints "iller".

The third and fourth string "filler" are really fillers as the name implies.

[nixen]

Thanks The pointer are the base of c++ language but in my life i don't use ** pointer, now i understand