printf / output questions in C

[Thiengineer]

Given the following code in C:

 
 char *s[] = {"program","test","load","frame","stack",NULL};
 char **p = s + 2;
 
 

Questions:

a) printf{"%s",p[-2] + 3);

b) printf("%c%s",* (s[1]),*p+1);

c) printf("%c%c%c% c", **s+1, *(*(p-1)+1), *(*s+3)+2, *(p[-1]+strlen(p[-1])-1));

d) *(*p+2) = '\0';
strncat(*p,p[2],2);
printf("%s",*p) ;


My Answers:

a) frame

b) toad

c) rea_ (the last character I don't know)

d) lost


Can someone verify if my answers are correct, if not please help me!

THANK YOU!!

[horace1]

I make the answers
(a) gram
(b) toad
(c) qeit

(d) causes a segmentation fault as the expression

 *(*p+2) = '\0';
 

attempts to change the 'a' in "load" to '\0' - "load" is a constant

[willakawill]

a) is gram because p[-2] is a char* pointing to the 'p' in 'program'. Incrementing this pointer by 3 brings us to 'g'. printf will print all characters until the null character so we get 'gram'
b) you are correct
c) **s is the character 'p' and p + 1 is 'q'
So we have 'q'

(p - 1) is 'test'
*(p - 1) is the char * pointing to 't'
(*(p - 1)+1) increments the pointer by one character to 'e'
*(*(p-1)+1) = 'e'
So we have 'qe'

*s is a char* pointing to 'p'
(*s + 3) increments this pointer to point to 'g'
*(*s + 3) is the char 'g'
'g' + 2 = 'i' therefore *(*s+3)+2 = 'i'
So we have 'qei'

strlen(p[-1]) is the length of 'test' which is 4
strlent(p[-1]) -1 is 4 - 1 = 3
p[-1] is a char* pointing to 't'
p[-1]+strlen(p[-1])-1 increments this char* by 3 to point to the last char in 'test' which is also 't'
*(p[-1]+strlen(p[-1])-1) is the character 't'
So we have 'qeit'

[Thiengineer]

Thank you for all your help!

I now understand a-c, but still uncertain about part d.

I don't understand how there is a segmentation fault, if that is the case.

Can someone please explain what's going on on part d?

Thanks again!

[willakawill]

You were correct in theory about the answer being 'lost' which is why I did not mention it. However, as Horace1 has pointed out, the code for part d raises an error. Perhaps you can get extra credits for pointing this out :)