String Pointer Tracing

[bobby19]

Hey. I am having trouble tracing this code. The output is

hursday March 30, 2006
March 30

but I cannot see why? From what I see, the str_ptr variable points at the String str, and this should point to the first element (T). Then the printf statement should print the letter h because it is incremented. After that it should print elemtent 8, which is y.

What am I missing here?

Thanks in advance

#include<stdio.h>
int main(void) {
char str[] = "Thursday March 30, 2006", *str_ptr;
str_ptr = str;
printf("%s\n", ++str_ptr);
str[17] = '\0';
printf("%s\n", str_ptr+7);
return(0);
}

[cwr]

str[8] is ' ', not 'y'.

Remember it starts with str[0].

[bobby19]

oops...but even so, why does it print the entire string after until the null character rather than the character. Would the pointer not only point to a specifc element?

[qqqqxxxx]

>printf("%s\n", ++str_ptr);
%c for character
and something like this *(str_ptr)

[vinit]

printf("%s\n", ++str_ptr);

You are making ++str_ptr so it will increment pointer first.
So its pre-increment,so first str_ptr is incremented then it is passed to printf() thats why ur seeing that output.

hursday March 30, 2006
March 30

Try

printf("%s\n", str_ptr++);

then U'll see result as

Thursday March 30, 2006
March 30

[bobby19]

So with these lines of code,

str_ptr = str;
printf("%s\n", ++str_ptr);
str[17] = '\0';
printf("%s\n", str_ptr+7);

am I essentially saying print the string from the place which str_ptr is pointing to until the NULL character is reached? Also why would not a defeference * be needed before the ++str_ptr and str_ptr+7?

[Salem]

Yes.

And it's called the nul character (\0)
This is different from the NULL pointer.

[jafet]

Exactly. That's how prinf() handles char*, and that's how all standard C handles char*, so you'd be better off doing it their way. Remember, a char* is simply a 32(or 64) bit pointer to somewhere in North Dakota.

--beaten by salem--


Note that even if your char[] points to "abcdefghijklmnop \\0\\ qrstuvwxyz", when passed to prinf() only "abcdefghijklmnop" will be printed. printf() makes no guesses regarding the actual size of your array, because there is no way to determine this at runtime.

[vinit]

Hi,bobby19

Also why would not a defeference * be needed before the ++str_ptr and str_ptr+7?

We don't dereference * b4 the ++str_ptr or str_ptr+7 is because we are just incrementing pointer not value inside it.
* is value of operator. so its not required to dereference * ++str_ptr.